There seems to be a weakness to the approach with your (generalized) differential equation for $v$, because the Laplacian of the indicator function $\Delta\mathbf{1}_{\partial\Omega}$ is not actually singular enough to produce a meaningful boundary condition on the surface $\partial\Omega$. We have that $\Delta\mathbf{1}_{\partial\Omega}=0$ wherever it is defined, and where it is undefined (that is, on $\partial\Omega$) it is merely undefined, not singular like a $\delta$-distribution (or derivative thereof). You can see this by integrating $\Delta\mathbf{1}_{\partial\Omega}$ times a test function and applying Cauchy's second formula (which is integration by parts twice), $$\int_{\mathcal{V}}dV\, \phi\left(\Delta\mathbf{1}_{\partial\Omega}\right)= \int_{\mathcal{V}}dV\, (\Delta\phi)\mathbf{1}_{\partial\Omega} +\int_{\partial\mathcal{V}}dA\left(\phi\frac{\partial\mathbf{1}_{\partial\Omega}}{dn}-\mathbf{1}_{\partial\Omega}\frac{\partial\phi}{\partial n} \right)=0,$$ where $\mathcal{V}$ is a larger volume containing $\Omega$. The integral is exactly the same as if $\mathbf{1}_{\partial\Omega}$ were replaced by $0$.

If the source term on the right-hand-side of the Poisson equation contained a singular distribution, that could yield a condition on a discontinuity of $v$ or its derivatives. However, with only $\Delta\mathbf{1}_{\partial\Omega}$, we are left with merely, $$\Delta v=0\quad \textrm{almost everywhere},$$ to which the generalized solution satisfying the boundary condition that $v|_{\partial\Omega}=0$ is just $v(x)=0$ everywhere. This is not actually giving a wrong answer; it just corresponds to the fact that the original boundary value problem for $u$ has the unique solution $u(x)=1$ on $\Omega$. However, your revised source is not actually contributing anything to the solution, so it is questionable whether this approach can be usefully generalized.

There seems to be a weakness to the approach with your (generalized) differential equation for $v$, because the Laplacian of the indicator function $\Delta\mathbf{1}_{\partial\Omega}$ is not actually singular enough to produce a meaningful boundary condition on the surface $\partial\Omega$. We have that $\Delta\mathbf{1}_{\partial\Omega}=0$ wherever it is defined, and where it is undefined (that is, on $\partial\Omega$) it is merely undefined (with a removable singularity), not singular like a $\delta$-distribution (or derivative thereof). You can see this by integrating $\Delta\mathbf{1}_{\partial\Omega}$ times a test function and applying Cauchy's second formula (which is integration by parts twice), $$\int_{\mathcal{V}}dV\, \phi\left(\Delta\mathbf{1}_{\partial\Omega}\right)= \int_{\mathcal{V}}dV\, (\Delta\phi)\mathbf{1}_{\partial\Omega} +\int_{\partial\mathcal{V}}dA\left(\phi\frac{\partial\mathbf{1}_{\partial\Omega}}{dn}-\mathbf{1}_{\partial\Omega}\frac{\partial\phi}{\partial n} \right)=0,$$ where $\mathcal{V}$ is a larger volume containing $\Omega$. The integral is exactly the same as if $\mathbf{1}_{\partial\Omega}$ were replaced by $0$.

If the source term on the right-hand-side of the Poisson equation contained a singular distribution, that could yield a condition on a discontinuity of $v$ or its derivatives. However, with only $\Delta\mathbf{1}_{\partial\Omega}$, we are left with merely, $$\Delta v=0\quad \textrm{almost everywhere},$$ to which the generalized solution satisfying the boundary condition that $v|_{\partial\Omega}=0$ is just $v(x)=0$ everywhere. This is not actually giving a wrong answer; it just corresponds to the fact that the original boundary value problem for $u$ has the unique solution $u(x)=1$ on $\Omega$. However, your revised source is not actually contributing anything to the solution, so it is questionable whether this approach can be usefully generalized.