Does invertability and closure imply identity?

Because if we were not essuming the existence of the identity element $e$, we would not be able to express the idea that $a^{-1}a=e$.

It's a reasonable question whether identity needs to be its own axiom. The inverse axiom requires that $a^{-1}$ exist for each $a$ such that $a^{-1} a=a a^{-1} = e$, which of course uses the identity in the definition. Instead, you might extend the inverse axiom as follows:

For each $a\in G$, there exists an element $a^{-1}\in G$, such that $a^{-1}a=aa^{-1}$ (call the result $e_a$), and such that for all $b\in G$, $e_a b=be_a =b$.

This seems to be weaker, allowing multiple distinct identity-ish elements, but it's not hard to show that they must actually all be the same element. So extending the inverse axiom in this way gets you back to the same definition of a group, except for one thing: without the identity axiom, the empty set would vacuously constitute a group.