You can satisfactorily explain Pascal's law, using the Euler equation for incompressible fluid flow. We can rewrite the equations in the following form
$$\rho \frac{\mathrm d\mathbf v}{\mathrm dt}=\mathbf f - \nabla p\tag{1}$$
where $\rho$ is the density of the fluid, $p$ is the pressure (scalar) function and $\mathbf f$ is the volume density of mass forces i.e. $\mathrm d\mathbf F/\mathrm dV$, where $\mathrm d\mathbf F$ is the net external force acting on the infinitesimal element inder consideration, and $\mathrm dV$ is its volume. In most of the cases, gravity is only the external force acting on the fluid, thus $\mathbf f$ becomes $\rho\mathbf g$.
Now, initially, let's assume the pressure varies as the function of the location of the infitesimal element. Thus we can represent it as $p(\mathbf r)$, where $\mathbf r$ is the position vector of the element. Now since the state is steady, thus $\mathrm d\mathbf v/\mathrm dt=0$. Thus, using equation $(1)$ we get
$$\nabla p(\mathbf r)=\rho \mathbf g\tag{2}$$
Now, let's say that the pressure function changes to another function, $p'(\mathbf r)$. Rewriting equation $(1)$, we get
$$\nabla p'(\mathbf r)=\rho \mathbf g\tag{3}$$
Comparing equation $(2)$ with equation $(3)$, we get
\begin{align}
\nabla p(\mathbf r)&=\nabla p'(\mathbf r)\\
\nabla \big(p(\mathbf r)-p'(\mathbf r)\big)&=0\tag{for all \(\mathbf r\))(4}
\end{align}
The equation $(4)$ can be true only when
$$p(\mathbf r)-p'(\mathbf r)=\text{constant}=\Delta p\tag{for all \(\mathbf r\)}$$
Thus the new pressure function must have increased by the same value everywhere. This is equivalent to saying that the pressure was transmitted everywhere equally.
