Using Laplace transform: $$\mathcal{L}_t[J_0(a \sinh (b t))](s)=\frac{I_{\frac{s}{2 b}}\left(\frac{a}{2}\right) K_{-\frac{s}{2 b}}\left(\frac{a}{2}\right)}{b}$$ where I'm found fromin this Book on page 276 example 1.
Using identity:
$$a \sin (b t)=-i a \sinh (i b t)$$
all combining together:
(-I)*(BesselI[s/(2 b), a/2] BesselK[-(s/(2 b)), a/2])/b /. a -> -I*2*x /. b -> I/2 /. s -> I*k -2 BesselI[k, -I x] BesselK[-k, -I x]
Sum[BesselJ[n, x]^2/(n - k), {n, -Infinity, Infinity}] == Re[-2 BesselI[k, -I x] BesselK[-k, -I x]] $$\sum _{n=-\infty }^{\infty } \frac{J_n(x){}^2}{n-k}=\Re(-2 I_k(-i x) K_{-k}(-i x))$$
Comment:
Formula works for:$x\in \mathbb{R}$ and $k=c+\frac{1}{2}$ where $c\in \mathbb{Z}$