It was a lot of fun for me! Here are the steps without much detail. It is a good exercise to check your understanding by supplying those details.
Let $\alpha + \beta = x$ and $\alpha - \beta = y$. The given equation is $ e^{i(x-\pi/2)}+e^{i(-y-\pi/2)}+e^{i(y+\pi/2)}+e^{i(-x+\pi/2)} = 0 $ which can be written as $ \begin{eqnarray*} e^{i(x-\pi/2)}+e^{i(-x+\pi/2)} &=& -e^{i(-y-\pi/2)}-e^{i(y+\pi/2)} \\ -i e^{ix}+ie^{-ix} &=& ie^{-iy}-i e^{iy} \\ e^{ix}-e^{-ix} &=& e^{iy}- e^{-iy} \\ 2i \sin x &=& 2i \sin y \\ \sin x &=& \sin y \\ y &=& n \pi + (-1)^{n} x \quad (n \text{ is an integer}) \\ \alpha + \beta &=& n \pi + (-1)^{n} (\alpha - \beta). \\ &\vdots& \end{eqnarray*} $ As Michael's comment points out, you have not mentioned what you are solving for, so I am unable to proceed further. But it is not hard to massage the above solution into the desired form.
Update (based on Michael's comments below). We can simplify this expression based on whether $n$ is odd or even. Suppose $n = 2m$ is even. Then, we have $ \alpha + \beta = 2m \pi + \alpha - \beta \quad\implies\quad \beta = m \pi. $ On the other hand, if $n = 2m+1$ is odd, then $ \alpha + \beta = (2m+1) \pi - (\alpha - \beta) \quad\implies\quad \alpha = \left(m + \frac12 \right) \pi. $
So the complete set of solutions are: $ \left \{ ( \alpha, m \pi ) \mid \alpha \in \mathbb R, m \in \mathbb Z \right\} \bigcup \left \{ \left( \frac { (2m +1) \pi}{2}, \beta \right) \mid \beta \in \mathbb R, m \in \mathbb Z \right\} . $