From Herstein's $Topics\ in\ Algebra$ Chapter 2, we have the following problem(s):
If a set under a closed associative product (binary) has:
$(1)$ An element $e\in G$ such that $\forall a\in G$, $ae=a$;
$(2)$ $\forall a\in G$, an element $a^{-1}$ such that $aa^{-1}=e$
then $G$ is a group. Further, if instead in $(2)$ we have instead that $a^{-1}a=e$, then prove, via example, that $G$ need not be a group.
$$$$I was able to prove the first part via manipulations, as follows (in short):
We have $e(a^{-1})^{-1}=aa^{-1}(a^{-1})^{-1}=ae=a\Rightarrow e(a^{-1})^{-1}=ea\Rightarrow ea=a$.
Further, we also get from the first equality that $a^{-1}aa^{-1}=a^{-1}\Rightarrow a^{-1}a=e$
A counterexample, for the second part is a set with the product defined as $xy=x$, for $x,y\in G$.
$$$$
My question is simply - I'm looking for some intuition behind why this holds, if indeed there is any available.