= Solution
For real $s>0$, collapse the Hankel contour onto the negative axis and use the jump in the argument of $t^{s-1}$. The reflection formula for the <gamma function> yields
$$
\boxed{
I(z,s)=\frac z{\Gamma(s)}
\int_0^\infty\frac{t^{s-1}}{e^t-z},dt},
$$
so $K(z,s,t)=z t^{s-1}/[\Gamma(s)(e^t-z)]$. At positive integer $s$, the original Hankel expression is interpreted by analytic continuation: the pole of $\Gamma(1-s)$ cancels the corresponding zero of the contour integral.
When $z$ crosses the slit at $x>1$, the pole of the real-integral kernel at $t=\log x$ crosses the integration path. The two boundary values differ by $2\pi i$ times its residue. Since $d(e^t-z)/dt=x$ there, that residue is $(\log x)^{s-1}/\Gamma(s)$, giving the jump
$$
\boxed{\frac{2\pi i(\log x)^{s-1}}{\Gamma(s)}}.
$$
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