I initially put it into the form of
\begin{align} X'' + \lambda^2 X &= 0 \\ Y'' - \lambda^2 Y &= 0 \end{align}
which has solutions
\begin{align} X &= a_n \cos(\lambda x) + b_n \sin(\lambda x) \\ Y &= c_n \cosh(\lambda y) + d_n \sinh(\lambda y) \end{align}
Given boundary conditions
\begin{align} y(0) &= U_{0} \\ x(0) &= 0 \\ x(a) &= 0 \end{align}
substituting into the solutions for $X, Y$ yields $a_n = 0$, $c_n = U_{0}$ and $\lambda = n \pi/a$.
My final solution is
$$u(x,y) = 2 U_{0} \sum_{n \ge 1} b_n \sin \left(\frac{n \pi x}{a} \right) e^{-n \pi y/a}$$
So it's close...but I am not sure how to get rid of the lambda in the exponential and the additional terms of $4$ and $\pi$
Your solution is incomplete. Separation of variables gives the general solution
$$ u(x,y) = \sum_{n=1}^\infty b_n \sin\left(\frac{n\pi x}{a}\right)e^{-n\pi y/a} $$
The boundary condition at $y=0$ requires
$$ u(x,0) = \sum_{n=1}^\infty b_n \sin\left(\frac{n\pi x}{a}\right) = u_0 $$
You need to find the Fourier series of the RHS to determine the $b_n$'s. This is given by
$$ b_n = \frac{2}{a}\int_0^a u_0 \sin\left(\frac{n\pi x}{a}\right) \ dx = \frac{2u_0}{n\pi}\big[1-\cos(n\pi)\big]$$
or
$$ b_n = \begin{cases} 0, & n \text{ even} \\ \dfrac{4u_0}{n\pi}, & n \text{ odd} \end{cases} $$
As for the missing constants in the exponents, I believe that's a typo in the text.