I concluded that $S$ converges ($x\in\mathbb R$, $a\notin\mathbb Z$): $$S=\displaystyle\sum_{k=-\infty}^\infty \dfrac{(-1)^k\exp ikx}{a^2-k^2}=\sum_{k=-\infty}^\infty \frac{(-1)^k \cos kx}{a^2-k^2}+i\displaystyle\sum_{k=-\infty}^\infty \dfrac{(-1)^k \sin kx}{a^2-k^2}.$$ Under the given conditions, $(-1)^k \cos kx\le 1$ and $(-1)^k \sin kx\le 1$, so if $\sum_{-\infty}^\infty \frac{1}{a^2-k^2}$ converges (which is true), then $S$ converges.
This result is contrary to Wolfram Alpha (it says the sum does not converge): https://www.wolframalpha.com/input/?i=sum+(-1)%5Ek+exp(ikx)%2F(a%5E2-k%5E2)+from+-inf+to+inf
The purported error perhaps shouldn't be the matter of $a$ not being in $\mathbb Z$: https://www.wolframalpha.com/input/?i=sum+1%2F(a%5E2-k%5E2)+from+-inf+to+inf
I don't know if I am missing something or if it's Alpha's error.