Show that a Modified-group action, is also a group action.

Question

Let $G$ be a group and $X$ a set. Define a modified-action as for which identity may not hold.

Have a subset $Y$ of the set $ X$ where holds:$\,e\star x\mid x.$

Can it be assumed that $e\star y$ has identity, as $e\star x= y, \implies e\star(e\star x) = e\star x$ doesn't help.
Unless that is possible, how to progress?

Answer 1

For closure, you need that given $e\star x\in Y$, that $g\star(e\star x)=e\star y$ for some $y\in G$.

So, let $y=g\star x$. Then $g\star (e\star x)=(ge)\star x=g\star x=(eg)\star x=e\star (g\star x)=e\star y$.