Making Fourier Transform $u(x,y)\to \hat{u}(k,y)$ we can rewrite the ODE as

$$\begin{cases}

-k^2\hat{u}+\hat{u}_{yy}=0\\

(\hat{u}_y+\alpha\hat{u})_{y=0}=\hat{f}(k)

\end{cases}$$

This ODE has solution $\hat{u}(k,y)=A(k)e^{-|k|y}+B(k)e^{|k|y}$. We will discard the second term because it goes unbounded as $y\to\infty$. Now using the boundary condition we have

$$(\hat{u}_y+\alpha\hat{u})_{y=0}=\hat{f}(k)\implies (-|k|+\alpha)A(k)=\hat{f}(k)\implies A(k)=\frac{\hat{f}(k)}{\alpha-|k|}$$

Now, if $\alpha<0$, then the RHS function has no singularities and we are good. To write the solution as fourier integral, we want IFT of $e^{-|k|y}/(\alpha-|k|)$, which we cannot find in closed form. So for now we can just write

$$u(x,y)=\frac{1}{2\pi}\int_{-\infty}^\infty f(x-z) \left(\int_{-\infty}^\infty \frac{e^{-|k|y+ikz}}{\alpha-|k|}\,dk\right)\,dz$$

If $\alpha>0$, then we have a problem as $\frac{\hat{f}(k)}{\alpha-|k|}$ could be singular. In this case, we need to require $\hat{f}(\alpha)=0$, which means

$$\int_{-\infty}^\infty f(x)e^{-i\alpha x}\,dx=0$$

This is a necessary condition, but not sufficient. $\hat{f}(k)$ needs to approach 0 faster than $\alpha-|k|$ does as $|k|\to\alpha$. Otherwise, we may still have problems with convergence. Assuming $\hat{f}(k)$ vanishes quickly enough at $\alpha$, the solution is given by the same equation.