I shall expand upon this post later with the full list of steps I took to cancel out all my eigenfunctions and eigenvalues. For the moment, my final answer was u=-1, which is bounded and satisfies all boundary conditions (trivially, $u_x$ and $u_y$ are 0 as needed).

Update (apologies for the comparatively short response, I'm not feeling great at the moment): so we first write $u=XY$ so that the problem becomes $\frac{Y''}{Y}+\frac{X''}{X}=0$. Condition (2) implies periodicity in x, so we have associated problems $X''+\lambda X = 0$, and $Y''-\lambda Y=0$. Then $X=A_n \cos(\omega_n x) + B_n \sin(\omega_n x)$, but we can set $B_n=0$ immediately from (2). Additionally, $X'(0)=X'(\frac{\pi}{2}=0) $ so $\omega_n=2n, \lambda_n=4n^2$. Then $Y = A_n e^{-2ny} + B_n e^{2ny} = A_n e^{-2ny} + 0$ , because u is bounded on $y>0$

So $u=A_0 + \sum_{n=1}^\infty e^{-2ny} \cos (2nx) \rightarrow (u_y-u)(x,0)=-A_0 - \sum_{n=1}^\infty (2n+1) A_n \cos (2nx) = 1 \rightarrow A_n(2n+1) = \frac{4}{\pi}\int_0^\frac{\pi}{2}\cos(2nx)$

In fact, this integral is 0 for all n. Therefore, we are just left with the condition $-A_0=1 \rightarrow u=-1$