To start with, given the metric ansatz
$$
ds^2 = e^{2\gamma(u)}dt^2 - e^{2\alpha(u)}du^2 - e^{2\beta(u)}d\Omega^2,\tag{1}
$$
one can apply the Cartan-Karlhede algorithm to check local equivalence with the Schwarzhschild solution (this is best carried out with a computer implementation). Instead of this general solution for the local equivalence of some metric ansatz and a given metric, one can in this case also investigate the demands for (1) to be a vacuum solution, since this will make it the Schwarzschild solution by virtue of being static and spherically symmetric (Birkhoff's theorem); the latter approach should of course also be used if one is primarily interested in the requirements for an ansatz to give a solution rather than some specific soluion.
To prove that $f(\alpha,\beta,\gamma) = 0$ is a "gauge" you can plug it into your ansatz and take the approach outlined above. For example for the ansatz (1) any "gauge" that does not violate the equations below is allowed.
I am new to gauge theory, but I believe the correct use of gauge transformations in general relativity would entail $SO_0(1,3)$-transformations of vectors in a rigid frame. Thus I believe you are correct in that this is a matter of coordinate freedom rather than gauge freedom.
Here I approach the metric ansatz (1) as an example: to that end I will be working in a local Lorentz (orthonormal) frame:
\begin{align}\begin{split}
\omega^0 &= e^{\gamma(u)}dt, \\
\omega^1 &= e^{\alpha(u)}du, \\
\omega^2 &= e^{\beta(u)}d\vartheta, \\
\omega^3 &= e^{\beta(u)}\sin(\vartheta) d\varphi.
\end{split}\tag{2}
\end{align}
We then work with the first Cartan equation
$$
d\omega^i = \gamma^i{}_{jk}\omega^j \wedge \omega^k,\tag{3}
$$
where $\gamma^i{}_{jk}$ are the Ricci rotation coefficients (components of the connection forms). Applying (3) to (2) we find
\begin{align}
\gamma^0{}_{jk}\omega^j\wedge\omega^k &= \gamma'(u)e^{\gamma(u)}du \wedge dt \\
&= \gamma'(u)e^{-\alpha(u)}\omega^1 \wedge \omega^0, \\
\gamma^1{}_{jk}\omega^j\wedge\omega^k &= 0, \\
\gamma^2{}_{jk}\omega^j\wedge\omega^k &= \beta'(u)e^{\beta(u)}du\wedge d\vartheta \\
&= \beta'(u)e^{-\alpha(u)}\omega^1 \wedge \omega^2, \\
\gamma^3{}_{jk}\omega^j\wedge\omega^k &= \beta'(u)e^{\beta(u)}\sin(\vartheta)du\wedge d\varphi + e^{\beta(u)}\cos(\vartheta)d\vartheta\wedge d\varphi \\
&= \beta'(u)e^{-\alpha(u)}\omega^1 \wedge \omega^3 + \cot(\vartheta)e^{-\beta(u)}\omega^2 \wedge \omega^3,
\end{align}
and by inspection we find
\begin{align} \begin{split}
\gamma^0{}_{10} &= \gamma'(u)e^{-\alpha(u)}, \\
\gamma^2{}_{12} &= \beta'(u)e^{-\alpha(u)}, \\
\gamma^3{}_{13} &= \beta'(u)e^{-\alpha(u)}, \\
\gamma^3{}_{23} &= \cot(\vartheta)e^{-\beta(u)},
\end{split}\tag{4}\end{align}
to be the only non-zero coefficients (up to symmetries: $\gamma_{ijk} = \gamma_{[ij]k}$). Next the second Cartan equation
\begin{align}
\frac{1}{2}R^i{}_{jk\ell}\omega^k \wedge \omega^\ell &= d\gamma^i{}_j + \gamma^i{}_k \wedge \gamma^k{}_j,\tag{5} \\
&= \left(-\gamma^i{}_{j[k|\ell]} + \gamma^i{}_{jm}\gamma^m{}_{[k\ell]} - \gamma_{mj[k}\gamma^{im}{}_{\ell]}\right)\omega^k \wedge \omega^\ell
\end{align}
gives, using (4):
\begin{align}\begin{split}
R^0{}_{101} &= \left(\alpha'(u)\gamma'(u) - \gamma''(u) - \gamma'(u)^2\right)e^{-2\alpha(u)}, \\
R^0{}_{202} &= -\beta'(u)\gamma'(u)e^{-2\alpha(u)}, \\
R^0{}_{303} &= -\beta'(u)\gamma'(u)e^{-2\alpha(u)}, \\
R^1{}_{212} &= \left(\alpha'(u)\beta'(u) - \beta''(u) - \beta'(u)^2\right)e^{-2\alpha(u)}, \\
R^1{}_{313} &= \left(\alpha'(u)\beta'(u) - \beta''(u) - \beta'(u)^2\right)e^{-2\alpha(u)}, \\
R^2{}_{323} &= e^{-2\beta(u)} - \beta'(u)^2e^{-2\alpha(u)},
\end{split}\tag{6}\end{align}
as the only non-zero Riemann components, again up to symmetries ($R_{ijk\ell} = R_{([ij][k\ell])}$), or in terms of the Ricci components (suppressing function arguments):
\begin{align}\begin{split}
R_{00} &= \left(2\beta'\gamma' - \alpha'\gamma' + \gamma'' + \gamma'^2 \right)e^{-2\alpha}, \\
R_{11} &= \left(\alpha'\gamma' + 2\alpha'\beta' - \gamma'' - 2\beta'' - \gamma'^2 - 2\beta'^2 \right)e^{-2\alpha}, \\
R_{22} &= \left(\alpha'\beta' - \beta' \gamma' - \beta'' - 2\beta'^2 \right)e^{-2\alpha} + e^{-2\beta}, \\
R_{33} &= \left(\alpha'\beta' - \beta' \gamma' - \beta'' - 2\beta'^2 \right)e^{-2\alpha} + e^{-2\beta}, \\
\end{split}\tag{7}\end{align}
The Ricci tensor must vanish in a vacuum solution (assuming no cosmological constant for simplicity), so (7) gives us a system of three differential equations:
\begin{align}
2\beta'\gamma' - \alpha'\gamma' + \gamma'^2 + \gamma'' &= 0, \tag{A}\\
\alpha'\gamma' + 2\alpha'\beta' - \gamma'' - 2\beta'' - \gamma'^2 - 2\beta'^2 &= 0, \tag{B}\\
\alpha'\beta' - \beta' \gamma' - \beta'' - 2\beta'^2 &= -e^{2(\alpha - \beta)}\tag{C}.
\end{align}
Notice that equations (A), (B), and (C) can be simplified:
If $\gamma' = 0$ we end up with a single equation:
$$
e^{\alpha} = \beta'e^\beta,
$$
under which all Riemann components (6) vanish. So this corresponds to flat (Minkowski) space.
Otherwise, by solving for $\gamma''$ in (A) and (B) we find $-\beta'\gamma' = \alpha'\beta' - \beta'' - \beta'^2$, allowing us to write
\begin{align}
\frac{\beta''}{\beta'} &= \alpha' - \beta' + \gamma', \tag{A*}\\
\frac{\gamma''}{\gamma'} &= \alpha' - 2\beta' - \gamma', \tag{B*}\\
2\beta'\gamma' + \beta'^2 &= e^{2(\alpha - \beta)},\tag{C*}
\end{align}
because $\beta' \neq 0$ by necessity, and we can integrate (A*) and (B*) to find
\begin{align}\begin{split}
\beta' &= C_1e^{\alpha - \beta + \gamma}, \\
\gamma' &= C_2e^{\alpha - 2\beta - \gamma},
\end{split}\tag{D}\end{align}
under which (C*) becomes
$$
C_1e^{-\alpha + \beta + \gamma}\left(2C_2e^{\alpha - 2\beta - \gamma} + C_1e^{\alpha - \beta + \gamma}\right) = 1,
$$
or
$$
e^{2\gamma} = C_1^{-2}\left(1 - 2C_1C_2e^{-\beta}\right)\tag{E1}.
$$
Note that (D) has already given us:
$$
e^{2\alpha} = C_1^{-2}\beta'^2e^{2(\beta - \gamma)}\tag{E2}.
$$
Equations (E1) and (E2) allows us to write (1) as
$$
ds^2 = \left(1 - 2Me^{-\beta(u)}\right)dt^2 - e^{2\beta(u)}\left(\frac{\beta'(u)^2}{1 - 2Me^{-\beta(u)}}du^2 + d\Omega^2\right),\tag{1.1}
$$
both by rescaling coordinates by constant factors if necessary. Notice that we have performed no (non-trivial) coordinate change. Thus (1.1) is just (1) rewritten to conform with the EFEs, excluding flat space, with an aptly chosen name on the integration constant.
We observe that imposing $\alpha(u) = -\gamma(u)$ is equivalent to restricting to $\beta' = e^{-\beta}$ or equivalently $\beta = \log(u + C)$ or $e^{2\beta} = (Cu)^{2}$ (new $C$), by (E2), which is just the Schwarzschild coordinates for (1.1). Similarly, from (B*) we can see that imposing $\alpha(u) = 2\beta(u) + \gamma(u)$ is equivalent to imposing $\gamma = Au + B$, which allows us to rewite (1.1) as (barring any calculation errors)
$$
ds^2 = e^{2Au}dt^2 - \frac{1}{\left(1 - C^2e^{2Au}\right)^{2}} \left(\frac{e^{2Au}}{\left(1 - C^2e^{2Au}\right)^2}du^2 + d\Omega^2\right).
$$
For example $\alpha(u) = -\gamma(u) + \beta(u)^2$ can similarly be plugged into e.g. (E2) to yield
$$
\int C_1^2 du = \int e^{2\beta - \beta^2} d\beta .
$$
As another example the condition $\beta = 2\gamma$ can be seen to demand a constant $\beta$, from (E1) and so cannot be used.
To be more general, any condition on the form $\alpha = f(\beta,\gamma)$ gives a first order ODE for $\beta$, by (E1) and (E2). By the standard existence theorem such a condition is consistent with the EFEs. On the other hand a condition on the form $\beta = f(\gamma)$ gives an equation for $\beta$, and so is not consistent with the EFEs. The consistency of imposing more than one condition can also be checked with the above equations. For example $\alpha = \beta = \gamma$ can immediately be seen to not be consistent with the EFEs, by (D).
Also, in this link you posted, you are already assuming Schwarzschild coordinate system a priori...
– Edison Santos Sep 16 '16 at 12:43