Corollary

A topological group $G$ is Hausdorff if and only if the intersection of the neighbourhoods of $e$ consists only of the point $e$.

The condition is clearly necessary. Conversely, if the intersection of all the neighbourhoods of $e$ is just $\{e\}$, then given any $x \neq e$ there is a neighbourhood $V$ of $e$ such that $x^{-1} \notin V$ and therefore $e \notin xV$. This shows that $x$ is not in the closure of $\{e\}$, and thus $\{e\}$ is closed, so that $G$ is Hausdorff.