So as you can immediately see, I have 0 at my corners where I should instead have $u(0,H) = u(L,0) = 1$ and $u(L,H) = 2$.
My problem was Laplace's Equation such that:
$$u_{xx} + u_{yy} = 0 \\ \cases{u(0,y) = y \\ u(L,y) = L+y \\ u(x,0) = x \\ u(x,H) = H+x} \\ u(x,y) = x+y$$
where $H=L = 1$.
Using the methodology from HERE on Example 6.3.2, I set up the problem by splitting it up into 4 different Laplace's Equations with 1 inhomogeneous condition from each boundary given to the 4 equations.
The general form of the solutions (for homogeneity along x) were $$u(x,y) = \sum_n{b_n\sin(\sqrt{\lambda}x)Y_n(y)} \\ b_n = \frac{2}{\sinh(\sqrt{\lambda}H)}\int_0^L f(x)dx \\ \lambda = (\frac{n\pi}{L})^2$$
I can immediately tell what the problem is, and that's the homogeneity of $\phi_n$ for $\lambda = \frac{n^2 \pi^2}{L^2}$ is causing 0 at x = 0, x=L (and this problem carries over to the homogeneity along y for y = 0 and y = H). If I were to take a guess, my solution for just one of the edges should look something like:
$$u(x,y) = b_0 + \sum_n{b_n\sin(\sqrt{\lambda}x)Y_n(y)} + b_L$$
and something similar to this for the other 3 edges.