The below theorem gives three conditions that form alternative group postulates.  It is not hard to show that they hold in the group defined ordinarily.

Proof.  If $a$ and $b$ are arbitrary elements, then there are at least one such $e_{a}$ and such $e_{b}$ that  $e_{a}a=a$  and  $be_{b}=b$.  There are also such $x$ and $y$ that  $xb=e_{a}$  and  $ay=e_{b}$.  Thus we have

i.e. there is a unique neutral element $e$ in $G$.  Moreover, for any element $a$ there is at least one couple $a^{{\prime}}$, $a^{{\prime\prime}}$ such that  $a^{{\prime}}a=aa^{{\prime\prime}}=e$.  We then see that

i.e. $a$ has a unique neutralizing element $a^{{\prime}}$.