A closed subset of an algebraic group which contains $e$ and is closed under taking products is a subgroup of $G$.
Denote this set as $X$. If the condition of $X$ being closed is dropped, this statement does not hold. The set of nonzero integers in $\mathbb{G}_m$ over $\mathbb{C}$ is a counterexample.
It suffice to prove that for any $x \in X$, $x^{-1}X = X$. As $X$ is closed under taking products, it is clear that $X \subseteq x^{-1}X$. In order to prove the inverse inclusion, the closedness of $X$ (under Zariski topology) must be used.
Let $\phi: G \rightarrow G, y \mapsto x^{-1}y$ is a homoemorphism of $G$ as an algebraic variety. So $x^{-1}X = \phi(X)$ is a closed subset of $G$ containing $X$. But Why are they equal?
Thanks very much.