Let $u=1/y^2$ and
$$f(u)=\int_{-\infty}^\infty{s\over(s-1)^2+1}\ln(1+u/s^2)\,ds$$
Note first that $s\ln(1+u/s^2)\to0$ both as $s\to0$ and as $s\to\pm\infty$, so the improper integral converges for all $u\ge0$, and, by dominated convergence, we have $\lim_{u\to0^+}f(u)=f(0)=0$. The limit we need to evaluate is $\lim_{u\to0^+}{f(u)\over u}$. L'Hopital tells us this is equal to $\lim_{u\to0^+}f'(u)$, provided that limit exists.
Working formally at first, we have
$$f'(u)=\int_{-\infty}^\infty{s\over(s-1)^2+1}\cdot{1\over s^2+u}\,ds$$
which also converges as long as $u$ is positive. (Note: If we let $u=0$ in this formula for $f'(u)$, the integrand has a pole at $s=0$ and the improper integral does not converge, unless one takes extra care to give it a "principal value" interpretation. But L'Hopital doesn't care about value of the derivative at $0$, just the values near $0$.)
Partial fractions lets us compute the indefinite integral:
$${s\over((s-1)^2+1)(s^2+u)}={1\over u^2+4}\left({(u-2)(s-1)+u+2\over(s-1)^2+1}-{(u-2)s+2u\over s^2+u}\right)$$
so that
$$\begin{align}
f'(s)
&={u-2\over u^2+4}\int_{-\infty}^\infty\left({s-1\over(s-1)^2+1}-{s\over s^2+u} \right)\,ds+{1\over u^2+4}\int_{-\infty}^\infty\left({u+2\over(s-1)^2+1}-{2u\over s^2+u} \right)\,ds\\\\
&={u-2\over u^2+4}\cdot{1\over2}\ln\left((s-1)^2+1\over s^2+u \right)\Big|_{-\infty}^\infty+{(u+2)\arctan(s-1)-2\sqrt u\arctan s\over u^2+4}\,\Big|_{-\infty}^\infty\\\\
&={(u+2-2\sqrt u)\pi\over u^2+4}
\end{align}$$
(in particular, the log term vanishes at $s=\pm\infty$), from which we see that
$$\lim_{u\to0^+}f'(u)={(0+2-2\sqrt0)\pi\over0^2+4}={\pi\over2}$$
and we are thus done, provided we justify the formalism of differentiating inside the integral. But this also comes courtesy of dominated convergence, since for any fixed positive value of $u$ and any appropriate small value of $h$ (so that $u+h$ is still positive), we have
$${f(u+h)-f(u)\over h}={1\over h}\int_{-\infty}^\infty{s\over(s-1)^2+1}\ln\left(1+{h\over s^2+u} \right)\,ds$$
and
$${1\over h}\left|{s\over(s-1)^2+1}\ln\left(1+{h\over s^2+u} \right) \right|\le{s\over((s-1)^2+1)(s^2+u)}$$
which, for any $u\gt0$, is integrable over $\mathbb{R}$. This lets us take the limit as $h\to0$ inside the integral sign, obtaining the asserted integral expression for $f'(u)$.