$(G,\mathcal T)$ is a semitopological group with $$(\forall x\in G)(\forall U\in \mathcal T)(\exists V \text{ neighborhood of }1)(x\notin U^cV)$$ Then $G$ is quasitopological (ie inverse is continuous).
Any idea to prove it or to find missing minimal necessary assumptions.
It seems the following.
To satisfy your condition, $x$ should be not in $U^c$, that is $x\in U$. Now if we take into account this correction, we obtain a corollary for $x=1$: for each neighborhood $U$ of the unit there exists a each neighborhood $U$ of the unit such that $1\not\in U^cV$, that is $V^{-1}\subset U$, which is equivalent to the continuity of the inversion.