The accepted answer for this question proves the following statement:
If $S$ is a closed subset of an algebraic group $G$ which contains $e$ and is closed under taking products in $G$, then $S$ is a subgroup.
I'd be interested to see an example of a closed subset of an algebraic group $G$ which is closed under products but does not contain $e$. Thanks!
I believe this is not possible, except for empty set. I think the argument used in the question you linked actually doesn't even use the hypothesis that $e\in S$. You can also reason as follows.
Consider a set $S_0$ with the properties you suggest and is nonempty. $\{e\}$ is closed (because Zariski topology is $T_1$), do $S=S_0\cup \{e\}$ is closed (as union of two closed sets), closed under products (clearly) and contains $e$, so by the cited fact it is a subgroup. But that implies that $S_0=S_0^{-1}$, and since it is nonempty and closed under products, $e\in S_0$.