Instead of asking whether there is a unified paradigm-spanning characterization of mass, you should first be asking what one would look like, if there were.
In Newton's original formulation - indeed in his very first two definitions - he defined the mass $m$ of a body to be what we in modern language express as the integral of the body's density over the region $V$ space covered by the body, and the momentum $$ of the body as the integral of the product of the density of each part of the body with the velocity which this part is moving at:
$$m = \int_V ρ() d^3,\quad = \int_V ρ() () d^3.$$
Provided that the density is non-negative everywhere $ρ() ≥ 0$ (and positive in the region, i.e. $ρ() > 0$ if $ ∈ V$), then you could also define the body's center of mass motion by assigning to it a velocity:
$$ ≡ \frac{\int_V ρ() () d^3}{\int_V ρ() d^3},$$
expressed as a density-weighted average of the localized velocity of its parts. With this, one can express the body's momentum in terms of its center of mass velocity as:
$$ = m .$$
The key properties of these quantities relate to how they transform under a change to a moving frame, i.e. under a "boost" transform. Suppose you change to a frame that is moving at a velocity $Δ$. Then, in the moving frame, the values of the density and localized velocities would respectively be transformed as:
$$ρ → ρ,\quad () → () - Δ(),$$
which yields the corresponding transforms for the mass, center of mass velocity and momentum:
$$m → m,\quad → - Δ,\quad → - m Δ.$$
The key take-home is the combined transform:
$$(m,) → (m, - m Δ).$$
Mass and momentum transform together, in a self-contained way, as a 4-vector.
In years following Newton, it was gradually realized that the kinetic (+ internal) energy, which will be denoted here as $H$, played in important role in dynamics. Expressed as:
$$H = ½ m ||^2 + U,$$
it consists of a part $½ m ||^2$ that is dependent on the body's center of mass motion, and a part $U$, that expresses the total (kinetic + internal) energy of the body's parts. The latter accounts for the motions of the body's parts relative to the body's center of mass, and for the internal energies in each of the parts. Being fixed to the body's center of mass, it is independent of the observer's frame of reference, so that one has the following transform under a boost:
$$H → ½ m | - Δ|^2 + U = H - ·Δ + ½ m |Δ|^2.$$
Combined with the mass and momentum, yields the following as a self-contained transform:
$$(m,,H) → \left(m, - m Δ, H - ·Δ + ½ m |Δ|^2\right).$$
This can be expressed more succinctly as an infinitesimal transform with an infinitesimal boost velocity $$ in place of $Δ$, keeping only quantities to the first order in $$, as the following set of infinitesimal changes:
$$δm = 0,\quad δ = -m ,\quad δH = -·.$$
It's actually easier to use transforms in infinitesimal form since they are linear. There's no real loss in doing so, since the finite version of the transform can be obtained by way of a Taylor series:
$$X + ΔX = X + s δX + \frac{s^2}{2!} δ^2X + \frac{s^3}{3!} δ^3X + ⋯,$$ with:
$$\begin{align}
δ(m,,H) &= (0,-m ,-·),\\
δ^2(m,,H) &= (0,-δm ,-δ·) = (0,,m||^2),\\
δ^3(m,,H) &= (0,,δm||^2) = (0,,0),\\
δ^n(m,,H) &= (0,,0)\quad (n > 3).
\end{align}$$
by setting $Δ = s$. Thus,
$$\begin{align}
(m,,H) + Δ(m,,H)
&= (m,,H) + s (0,-m ,-·) + \frac{s^2}{2} (0,,m||^2)\\
&= \left(m, - ms, H - ·s + ½m||^2s^2\right)\\
&= \left(m, - m Δ, H - ·Δ + ½m|Δ|^2\right).
\end{align}$$
Now ... we get to Relativity, in which a new talking point is added: mass-energy conversion. If we use a similar expression for mass versus momentum, $ = M $, distinguishing it from the invariant mass $m$, then according to Relativity, there is an additional contribution to $M$ arising from the mass equivalent of the body's kinetic energy, $H$, so that one has the following decomposition:
$$M = m + \frac{H}{c^2}.$$
If we continue to assume $m$ is invariant (i.e. $δm = 0$), and that $H$ has the same infinitesimal transform ($δH = -·$), and that the same ratio $M$ applies both to the finite form of the expression for momentum ($ = M $) as does to the infinitesimal transform ($δ = -M $), then this entails the following set of infinitesimal transforms:
$$δ(M,,H) = \left(-\frac{·}{c^2}, -M, -·\right),\quad δm = 0,\quad δ = - + \frac{·}{c^2}.$$
The extra term in the transform of $$ is needed to ensure the consistency between $ = M $ and $δ = -M $, but it can also be connected to the Lorentz transform. If we write the differential equation for the velocity $ = d/dt$ in differential form and apply the transform separately to the coordinate differentials $d$ and $dt$, we get:
$$\begin{align}
&d = dt\\
&⇒\quad δ(d) = \left(- + \frac{·}{c^2}\right)dt + δ(dt)\\
&⇒\quad δ(d) + dt = \left(δ(dt) + \frac{·dt}{c^2}\right)\\
&⇒\quad δ(d) + dt = \left(δ(dt) + \frac{·d}{c^2}\right)
\end{align}$$
and assume that this holds for arbitrary $$, then it follows that:
$$
= δ(d) + dt\quad⇒\quad δ(d) = - dt,\\
0 = δ(dt) + \frac{·d}{c^2}\quad⇒\quad δ(dt) = -\frac{·d}{c^2}.
$$
This is the infinitesimal version of the Lorentz transform. You can run this sequence in reverse to get the transform for $$ and - from this - the connection between $ = M $ and $δ = -M $ that they both be with the same ratio $M$.
In Relativity, one normally combines the mass and kinetic energy into the total energy $E$, via the formula $E = M c^2$, as:
$$E = \left(m + \frac{H}{c^2}\right)c^2 = m c^2 + H.$$
This allows you to rewrite the transform of the five-vector $(M,,H)$ in split form as self-contained transforms
$$δ(E,) = \left(δm c^2 + δH, δ\right) = \left(-·, -M\right) = \left(-·, -\frac{E}{c^2}\right),\quad δm = 0,$$
split between the four-vector $(E,)$ and scalar $(m)$.
However...
there is no counterpart, in Relativity, to the "$U$" of the expression $H = ½m||^2 + U$, which means that the value of $H$ in the rest frame of the body is just 0, and correspondingly $m$ is just the rest frame value of $M$. So, it is no longer an independent quantity. Instead, one has the constraint:
$$\frac{E^2}{c^2} = ||^2 + m^2 c^2,$$
which you can prove by (a) showing that the equation reduces to an identity in the body's rest frame and (b) showing that both sides have the same infinitesimal transform.
If we were to go beyond Relativity and actually attach a counterpart to $U$, then we would need to draw a distinction between $m$ as "rest mass" versus $μ = m$ as "invariant mass", renaming the latter $μ$. In that case, the decomposition for $M$ and expression for $m$ would be rewritten, respectively, as
$$M = μ + \frac{H}{c^2},\quad m = μ + \frac{U}{c^2}$$
and with the transform $δμ = 0$.
Why the "beyond Relativity" part? Because it's only there that you will find the stitching together of the non-relativistic and Relativistic accounts for mass into the seamless unifying framework that you're asking for.
So, now the actual splitting of $(M,,H)$ is into $(E,)$ and $(μ)$, instead of into $(E,)$ and $(m)$. Instead, $m$ arises as the norm of $(E,)$:
$$(mc)^2 = |(E,)|^2 = \frac{E^2}{c^2} - ||^2,$$
rather than as an independent quantity in its own right.
In Relativity, the finite version of the transform $δ(E,)$, when applied in reverse to the rest frame values $(mc^2,)$ for a body yield their values in a moving frame:
$$(E,) = (Mc^2, M),\quad M^2 = m^2 + \frac{||^2}{c^2} = m^2 + M^2\frac{||^2}{c^2}\quad⇒\quad M = \frac{m}{\sqrt{1 - ||^2/c^2}}.$$
From this follows
$$H = (M - μ)c^2 = \left(E - mc^2\right) + U = \frac{M||^2}{1 + \sqrt{1 - ||^2/c^2}} + U.$$
So, what are these quantities and what is the unified account for them?
In Relativity, you can verify that $E dt - ·d$ is an invariant as follows:
$$\begin{align}
δ(E dt - ·d)
&= δE dt + E δ(dt) - δ·d - ·δ(d)\\
&= (-·) dt + E \left(-\frac{·d}{c^2}\right) - \left(-\frac{-E}{c^2}\right)·d - ·(-dt)\\
&= 0.
\end{align}$$
So, $(E,)$ transform as $(∂/∂t,-∇)$.
However, if you try this make this work with the non-relativistic version $H dt - ·d$ and - specifically - with the kinetic energy, you'll instead get:
$$\begin{align}
δ(H dt - ·d)
&= δH dt + H δ(dt) - δ·d - ·δ(d)\\
&= (-·) dt + H \left(-\frac{·d}{c^2}\right) - \left(-M\right)·d - ·(-dt)\\
&= \left(M - \frac{H}{c^2}\right)·d\\
&= μ·d.
\end{align}$$
The non-relativistic version of this derivation - obtained by setting all of the $1/c^2$ terms to $0$ - yields the same result.
To make this invariant requiring slipping in an extra coordinate $u$ whose differential transforms as
$$δu = ·d.$$
Then, we can write down the following invariant:
$$H dt - ·d - μ du = H ds - ·d - M du,\quad s = t + \frac{u}{c^2}.$$
In the process, you'll also find another invariant emerge:
$$δ(ds) = δ(dt) + \frac{δ(du)}{c^2} = -\frac{·d}{c^2} + \frac{·d}{c^2} = 0.$$
The coordinate $s$ is the "beyond" Relativistic version of the absolute time of non-relativistic theory, and the 5-vector $(H,,μ)$ transforms as $((∂/∂t)_u,-∇,-(∂/∂u)_t)$. Similarly, the 5-vector $(H,,M)$ transforms as $((∂/∂s)_u,-∇,-(∂/∂u)_s)$.
The 5-dimensional geometry with coordinates $(t,u,=(x,y,z))$ - in the non-relativistic case - is known as Bargmann geometry. It is characterized by the coordinate invariants:
$$|d|^2 + 2 dt du,\quad ds = dt.$$
The Relativistic version is a morphed version of Bargmann with the following as its coordinate invariants:
$$|d|^2 + 2 dt du + \frac{(du)^2}{c^2},\quad ds = dt + \frac{du}{c^2}.$$
A unified account for mass that spans the paradigm divide between Relativity and non-relativistic theory can only be found by first providing a unified account of their respective geometries, and only within it will you find your answer. Just as the total energy $E$ and momentum $$ - in Relativity - can be expressed as conjugate to $t$ and $$ (up to sign), the kinetic energy $H$, momentum $$ and invariant mass $μ$ can - in both the "beyond" Relativistic and non-relativistic settings - be expressed (again, up to sign) as conjugates respectively to the coordinates $t$, $$ and $u$. Similarly, $H$, $$ and $M$ can be treated as conjugate to $s$, $$ and $u$, where $s$ is taken as a coordinate in place of $t$.
In the unified account you're looking for, mass (either as $μ$ or $M$) is conjugate to $u$.
In dynamics given by an action integral $S = \int L(,t,u) ds$, requiring independence of $u$ leads - by way of the Noether theorem - to a conservation law for $μ$. If you drop down from "beyond" Relativity to just ordinary Relativity, where $μ = m$, or down to the non-relativistic setting, that leads to a conservation law for $m$. For a single free body, you can directly write it as a Lagrange multiplier:
$$S = \int \frac{m}{2} \left(\left|\frac{d}{ds}\right|^2 + 2\frac{dt}{ds}\frac{du}{ds} + α\left(\frac{du}{ds}\right)^2\right) ds = \int \frac{m}{2} \frac{\left|\frac{d}{dt}\right|^2 + 2\frac{du}{dt} + α\left(\frac{du}{dt}\right)^2}{1 + α\frac{du}{dt}}dt.
$$
In the Relativistic case ($α = 1/c^2$) or non-Relativistic case ($α = 0$) that yields the correct values for the conjugate momentum, after eliminating the $u$ coordinate via the equation
$$\left|\frac{d}{dt}\right|^2 + 2\frac{du}{dt} + α\left(\frac{du}{dt}\right)^2 = 0.$$
The Euler-Lagrange equation for $u$ will - after applying this constraint - just be
$$\frac{dm}{dt} = 0,$$
while for $$, it will be
$$\frac{d}{dt}\frac{m(d/dt)}{1 + α(d/dt)} = ,$$
or, since
$$\left(1 + α\frac{du}{dt}\right)^2 = 1 - α\left|\frac{d}{dt}\right|^2,$$
it can be written equivalently as
$$\frac{d}{dt}\frac{m(d/dt)}{\sqrt{1 - α|d/dt|^2}} = .$$
