= Solution
Take the <branch of the complex logarithm> with $-\pi<\arg s<\pi$ and write $s^{-1/2}=\exp[-\log s/2]$. Deform the <Bromwich contour> into a Hankel contour about the negative real axis. The small circle at zero contributes $O(r^{1/2})$ and vanishes; the large closing arcs vanish by the exponential estimates for $t>0$. On the upper lip $s=-r+i0$, $s^{-1/2}=-i/\sqrt r$, while on the lower lip it is $i/\sqrt r$. Keeping the lip orientations in the <Bromwich inversion with a square-root branch cut> gives
$$
f(t)=\frac1{2\pi i}\int_0^\infty e^{-rt}\left(\frac{i}{\sqrt r}-\frac{-i}{\sqrt r}\right)dr
=\frac1\pi\int_0^\infty r^{-1/2}e^{-rt}dr.
$$
With $r=u^2$, the supplied <Gaussian integral> evaluates this as $2\pi^{-1}\int_0^\infty e^{-tu^2}du$. Hence
$$
\boxed{\mathcal L^{-1}[s^{-1/2}](t)=\frac1{\sqrt{\pi t}}\quad(t>0).}
$$
The inverse is singular but locally integrable at zero, which is compatible with existence of its <Laplace transform>.
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