[Disclaimer: the below text is not mine, but by Vladimir Petrov, who does not have MO account]
Consider generalized Mehler-Fock transform:
$$
\begin{cases}
F(\xi,\,\mu)&=\intop_1^\infty f(y)P^{-\mu}_{-1/2+i\xi}(y)\,dy,\ \ \ 0\le\Re\mu<1;\\
f(x)&={1\over\pi}\intop_0^\infty \xi\sinh\pi\xi\,\Gamma(\mu+1/2+i\xi)\Gamma(\mu+1/2-i\xi)
P^{-\mu}_{-1/2+i\xi}(x)F(\xi,\,\mu)\,d\xi.\,\,(*)
\end{cases}
$$
We have
$$\intop_0^\infty f(\alpha)\,Q_{s-1}^{i\sqrt{\alpha}}(z)\,d\alpha=A(s,z).\,\,(1)$$
Change bariables to $\xi=\sqrt{\alpha}$ and use Wipple transform (Bateman, Erd\'elyi vol. 1, p. 3.3.1, formula (14)). We get equation
$$\sqrt{2\pi}\intop_0^\infty g(\xi)\xi e^{-\pi\xi}\,\Gamma(s+i\xi)\,(z^2-1)^{-1/4}\,P_{-1/2-i\xi}^{1/2-s}\left({z\over\sqrt{z^2-1}}\right)\,d\xi
=A(s,z).\,\,(2)$$
Now set $z/\sqrt{z^2-1}=x$. Then $z^2-1=(x-1)^{-1}.$ And the answer is (calculations have to be checked):
$$\sqrt{2\pi^3}\,g(\xi){e^{-\pi\xi}\over\sinh\pi\xi\,\Gamma(s-i\xi)}=\intop_1^\infty A\left(s,{x\over x-1}\right)\,(x-1)^{-1/4}P_{-1/2-i\xi}^{1/2-s}(x)\,dx \,\,(3)$$
Something should be said about analytical continuation on $s$ to the right half-plane. Transform $(*)$ is defined in the strip $0<\Re\mu<1,$
but the same formula holds for integer $\mu.$ Since Legendre function is analytical in upper index in the right
halfplane, Carlson theorem implies that the function holds in the whole right halfplane.
