Let $\left(A,\ *, T\right)$ be a Ring and $e$ be its identity element.
Is it possible to prove the following statement without using the regularity of $aTe$ in the underlying group $\left(A,\ *\right)$?
$$\left(\forall\ a \in A \right)\ aTe = eTa = e$$
Proof using regularity:
$$aTe = a T \left(e * e \right)$$
$$\Leftrightarrow aTe = \left(a T e \right)* \left(a T e \right)$$
$$\Leftrightarrow \left(aTe\right)*e = \left(a T e \right)* \left(a T e \right)$$
Since $aTe \in A$ and $\left(\forall\ x \in A \right)\ x$ is regular in $\left(A,\ *\right)$, then $e = \left(a T e \right)$
The same way, we can prove $eTa = e$
I appreciate any hints!