Take $\tilde{\Gamma}$ to be a higher genus Baumslag-Solitar group, see section 9 of M. Kapovich, B. Kleiner, Coarse Alexander duality and duality groups, Journal of Differential Geometry, Vol. 69, (2005) Number 2, p. 279-352. Group $\tilde\Gamma$ cannot act properly discontinuously on any contractible 3-manifold, but it contains a finite index subgroup $\Gamma$ which is is Gromov-hyperbolic and isomorphic to the fundamental group of a compact 3-manifold with boundary. Hence, by a corollary of Thurston's geometrization theorem, $\Gamma$ acts isometrically and discretely on hyperbolic 3-space, i.e., is a discrete subgroup of $G=PO(3,1)$.
Edit: Here is an easier example. Note first that every finite group admitting an effective topological action on $R^2$ is either cyclic or dihedral. Now, take, say, a simple noncyclic finite group $F$ and set $\tilde\Gamma= F * F$. Then $\tilde\Gamma$ is virtually free, so it contains a free finite index subgroup $\Gamma$, which embeds in $PSL(2, R)$. On the other hand, every topological $\tilde\Gamma$-action on the plane has to be trivial since each free factor has to act trivially.