Assuming that you enjoy special functions
$$\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}=\frac{-\psi \left(-x-\sqrt{-a}+1\right)+\psi
\left(-x+\sqrt{-a}+1\right)-\psi \left(x-\sqrt{-a}\right)+\psi
\left(x+\sqrt{-a}\right)}{2 \sqrt{-a}}$$
wich can simplify as
$$\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}=-\frac{\pi \left(\cot \left(\pi \left(\sqrt{-a}-x\right)\right)+\cot \left(\pi
\left(\sqrt{-a}+x\right)\right)\right)}{2 \sqrt{-a}}$$ and, since $a >0$
$$\color{blue}{\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}=\frac{\pi \left(\coth \left(\pi \left(\sqrt{a}-i x\right)\right)+\coth \left(\pi
\left(\sqrt{a}+i x\right)\right)\right)}{2 \sqrt{a}}}$$ what you can simplify using
$$\coth(A+B)+\coth(A-B)=\frac{2 \sin (2 A)}{\cos (2 B)-\cos (2 A)}$$
Edit
In order to keep the results in the answer, I reproduce here what you wrote in comments.
So, we have finally
$$\sum_{k=-\infty}^\infty\frac1{a+(k+x)^2}=\frac \pi {\sqrt a}\,\,\frac{\sinh \left(2 \pi \sqrt{a}\right)}{\cosh \left(2 \pi \sqrt{a}\right)-\cos(2 \pi x)}$$ but I think that you cannot make at the same time $a \to 0$ and $x\to \pm 1$ without avoiding $\infty$. In the post, remember that you did precise $x\in(-1,1)$ and not $x\in[-1,1]$.