I've found the below equivalence.
For $a,b,x\in\mathbb{C}$, provided that there are no singularities on the right-hand side:
\begin{multline}\sum _{k=2}^{\infty}\sum _{j=1}^{\infty}\frac{x^k}{(a j+b)^k}=-\frac{x^2}{2b(b-x)}+\frac{\pi x}{2a}\csc{\frac{b\pi }{a}}\csc{\frac{\pi (b-x)}{a}}\sin{\frac{\pi x}{a}}\\-\frac{x \pi}{a}\int _0^1\left(\csc{\frac{2 \pi (b-x)}{a}}\sin{\frac{2 \pi (b-x)u}{a}}-\csc{\frac{2 \pi b}{a}}\sin{\frac{2 \pi b u}{a}}\right)\cot{\pi u}\,du\end{multline}
Even when the left-hand side may not converge, the right-hand side may still converge.
I've notice that $b=1/2$ causes the integral (the imaginary part for real $a,b$) to vanish for $x=1$ and $a=-i$ (though the real part doesn't vanish).
The integral will always vanish for $x=2b$, so it's a matter of finding a triple $(a,b,2b)$ that zero out the other part.
What are the zeros of this equation?
For $|x| < |i\Im(b)+1+ \max(\Re(b),0)|$ and $b \not \in \mathbb{Z}_{\le 1}$ then
$$\sum_{k=2}^{\infty}\sum _{j=1}^{\infty}\frac{x^k}{( j+b)^k}=\sum _{j=1}^{\infty}\sum_{k=2}^{\infty}\frac{x^k}{( j+b)^k} = \sum_{j=1}^\infty \frac{1}{1-\frac{x}{j+b}}-1-\frac{x}{j+b}$$ $$= \sum_{j=1}^\infty \frac{x}{j+b-x}-\frac{x}{j+b}=x (\frac{\Gamma'(b+1)}{\Gamma(b+1)}-\frac{\Gamma'(b-x+1)}{\Gamma(b-x+1)})$$