Let $k > 0$ and $ a>1$ be constants. As far as I can tell, the integral $$ J = \int_{-\infty}^\infty dx\frac{e^{i k x}}{1+x^2}\frac{1}{\log(a - ix)} $$ converges, since the argument of the logarithm never equals $1$ and is never a nonpositive real number. Calculating it using a contour in the upper half-plane, I find that it equals $$ 2\pi i \frac{e^{-k}}{2i}\frac{1}{\log(a+1)}= \frac{\pi \,e^{-k}}{\log(a+1)}. $$
Clearly this expression is not valid if $k < 0$, because the integral should not blow up. For $k < 0$, we would like to use a contour in the lower half-plane. But the logarithm has a branch point $z = -i a$ there. Can we calculate the integral for $k < 0$ using complex analysis or otherwise?
Verification
Since the integrand vanishes in the extremes of the upper half-plane, close the path of integration in the upper half-plane. $$ \gamma=[-R,R]\cup Re^{i[0,\pi]} $$ The branch point of $\log(a-ix)$ is at $-ia$ which is outside $\gamma$.
The integral along the circular part of $\gamma$ vanishes as $R\to\infty$, leaving us with $$ \begin{align} \int_{-\infty}^\infty\frac{e^{ikx}}{1+x^2}\frac{\mathrm{d}x}{\log(a-ix)} &=2\pi i\operatorname*{Res}_{z=i}\left(\frac{e^{ikz}}{1+z^2}\frac1{\log(a-iz)}\right)\\ &=2\pi i\frac{e^{-k}}{2i}\frac1{\log(a+1)}\\ &=\frac{\pi e^{-k}}{\log(a+1)} \end{align} $$ So your integral appears correct.
Negative $\boldsymbol{k}$
We can do the same thing for negative $k$, except we need to add a slit along $(-ia,-i\infty)$ for the branch cut and include the residue from the singularity at $-i(a-1)$. If we parametrize the slit as $z=-ix$, then we have $\log(a-iz)=\log(x-a)+i\pi$ on the left of the slit and $\log(a-iz)=\log(x-a)-i\pi$ on the right. Then $$ \begin{align} &\overbrace{\int_{-\infty}^\infty\frac{e^{ikx}}{1+x^2}\frac{\mathrm{d}x}{\log(a-ix)}}^{\text{part along the real line ($z=x$)}} \overbrace{-i\int_a^\infty\frac{e^{kx}}{1-x^2}\left(\frac1{\log(x-a)+i\pi}-\frac1{\log(x-a)-i\pi}\right)\mathrm{d}x}^{\text{part around the slit ($z=-ix$)}}\\ &=\int_{-\infty}^\infty\frac{e^{ikx}}{1+x^2}\frac{\mathrm{d}x}{\log(a-ix)} +\int_a^\infty\frac{e^{kx}}{x^2-1}\frac{2\pi}{\log(x-a)^2+\pi^2}\mathrm{d}x\\ &=-2\pi i\operatorname*{Res}_{z=-i}\left(\frac{e^{ikz}}{1+z^2}\frac1{\log(a-iz)}\right)-2\pi i\operatorname*{Res}_{z=-i(a-1)}\left(\frac{e^{ikz}}{1+z^2}\frac1{\log(a-iz)}\right)\\ &=-2\pi i\frac{e^{k}}{-2i}\frac1{\log(a-1)}-2\pi i\frac{e^{k(a-1)}}{a(2-a)}i\\ &=\frac{\pi e^k}{\log(a-1)}+\frac{2\pi e^{k(a-1)}}{a(2-a)} \end{align} $$ Therefore, for $k\le0$, $$ \begin{align} &\int_{-\infty}^\infty\frac{e^{ikx}}{1+x^2}\frac{\mathrm{d}x}{\log(a-ix)}\\ &=\frac{\pi e^k}{\log(a-1)}+\frac{2\pi e^{k(a-1)}}{a(2-a)}-\int_a^\infty\frac{e^{kx}}{x^2-1}\frac{2\pi}{\log(x-a)^2+\pi^2}\mathrm{d}x \end{align} $$ When $k=0$, this equals $\frac\pi{\log(a+1)}$. However, for $k\lt0$, this does not equal $\frac{\pi e^{-k}}{\log(a+1)}$.