Thank to KCd's hint. Let me have a trying.
Proof.
It's harmless to suppose $U$ is open.
Since $G$ is topological group, $\cdot^{-1}[U]$ is open and contains $(e,e)$, hence there is a $A \times B$, in which $A$ and $B$ are both open, contains $(e,e)$ and $\subseteq \cdot^{-1}[U]$. Consequently $V \times V$ is an open neighborhood of $(e,e)$ too, in which $V=(A \cap B) \cap (A \cap B)^{-1}$. Note that $\cdot$ is an open mapping, hence $V \cdot V$ is open. Besides $e \in V \cdot V \subseteq U$ since $(e,e) \in V \times V \subseteq \cdot^{-1}[U]$. Moreover $V^{-1}=V$.
Therefore $V$ satisfies all the conditions required.
$\Box$