In a Hausdorff topological group, both multiplication and inversion maps are continuous, so inverse image of any open set is always open under multiplication as well as inversion. Thus to find a topological group in which either multiplication or inversion doesn't always pullback open sets to $F_\sigma $ sets, we must only consider non-metrizable groups as in metric spaces every open set is $F_\sigma$.
So if we consider any group with indiscreet topology, say $H$, and take box product of uncountable copies of it, i.e. $G=H^I$ (box topology), we will get a non-metric topological group. In this group $G$, if we look at inversion mapping, the pullback of element of $G$, where every element of "uncountable tuple" (not sure how to write it precisely) is $H$, must not be $F_\sigma$. I am not sure how convincing this example is, or if there are other better examples out there. Thanks.
Take any $G$ that is a topological group that is not perfectly normal: there is some open set $O$ that is not an $F_\sigma$. Then $i^{-1}[O]$ is also not an $F_\sigma$ as the inversion $i$ is a homeomorphism from $G$ to itself.
So an example is trivial, in a way. E.g. take the group $\{0,1\}^I$ (in the usual product topology) for an uncountable index set $I$, with coordinatewise addition mod $2$ as a group operation, as a standard example of such a group.