I am trying to solve:
$$\nabla^2 u(x,y) = \partial_{xx} u(x,y) + \partial_{yy} u(x,y) = 1-y$$
over the domain $[-1,1]\times [0,1]$, with the B.C.:
$$u(x,0) = 0 = u(\pm 1, y).$$
Is there any closed-form solution to that PDE?
Assuming the solution is polynomial, I could show that $u(x,y)=\tfrac{1}{2}(1-x^2)y$ is solution to $\nabla^2 u = -y$ + B.C. but that does not answer the question... If I'm not mistaken, there is no polynomial $u$ solution.
There is a variety of solutions that can be constructed, but none of them are polynomial. Since you already found a function that satisfies $\nabla^2 u(x,y)=-y$, let's attack the problem $$\nabla^2 u(x,y)=1\\ u(x,0)=0~~,~ u(\pm 1, y)=0$$
First we need to find a complete set of eigenfunctions of the Laplacian with these boundary conditions. To that end we perform separation of variables, which after applying BC's yields two infinite families of solutions
$$\{\sin n\pi x\sinh n\pi y ~,~~ n\geq 1 \},~\{\cos (n+1/2)\pi x\sinh (n+1/2)\pi y ~,~~ n\geq 0 \} $$
that satisfy the constraints. The two families are linearly independent of each other. We note that only the second family has overlap with the constant function in $x$, since $\int_{-1}^1 \sin(n\pi x)dx=0$. To find a particular solution to the equation, set
$$u(x,y)=\sum_{n\geq 0}f_n(y)\cos(n+1/2)\pi x+ \sum_{n\geq 1}g_n(y)\sin n\pi x $$
Plugging in to the PDE and using the fact that $$1=\frac{2}{\pi}\sum_{n=0}^\infty(-1)^n\frac{\cos(n+1/2)\pi x}{n+1/2}$$ we obtain the equation
$$\sum_{n\geq 0}(f''_n(y)-\pi^2(n+1/2)^2f_n(y))\cos(n+1/2)\pi x+\sum_{n\geq 1}(g_n''(y)-n^2\pi^2 g_n(y))\sin n\pi x= \frac{2}{\pi}\sum_{n=0}^\infty(-1)^n\frac{\cos(n+1/2)\pi x}{n+1/2}$$
which yields the following families of ODE's:
$$f_{n}''(y)-\pi^2(n+1/2)^2f_n(y)=\frac{2}{\pi(n+1/2)}(-1)^n~~,~ f_n(0)=0\\ g_n''(y)-(n\pi)^2g_n(y)=0~~,~ g_n(0)=0 $$
with general solutions
$$f_n(y)=A_n \sinh (n+1/2)\pi y+ 2\frac{(-1)^{n+1}}{\pi^3(n+1/2)^3}(e^{-\pi(n+1/2)y}-1)$$ $$g_n(y)=B_n \sinh n\pi y$$
The arbitrary constants $A_n, B_n$ represent homogeneous solutions, and we set them to zero to obtain a particular solution
$$u_p(x,y)=\sum_{n=0}^\infty\frac{16}{\pi^3(2n+1)^3}(-1)^n (e^{-\pi(n+1/2)y}-1)\cos\pi(n+1/2)x$$
This series can be expressed in terms of polylogarithms and polynomials, but never just polynomials.