A Left Identity Forces Equality with Unity

Detailed Solution

Since $1$ is the (two-sided) multiplicative identity of $R$, it satisfies $x\cdot1=x$ for every $x\in R$, in particular for $x=e$: $$ e\cdot1=e. $$ On the other hand, the hypothesis on $e$ says $e\cdot a=a$ for every $a\in R$, so taking $a=1$ gives $$ e\cdot1=1. $$ Combining these two equalities, $$ e=e\cdot1=1. $$ So $e=1$ is forced, and this argument used only the defining property $x\cdot1=x$ of the unity together with the hypothesis on $e$ — commutativity of $R$ was never needed. This rules out option B. Option C fails because if $e=0$, then $e\cdot a=0\neq a$ for any nonzero $a$ (recalling $1\neq0$ guarantees $R$ has nonzero elements), contradicting the hypothesis. Option D is also false, since the argument above works in any ring with unity, commutative or not, and does not require $R$ to be a field. Thus the correct option is $\boxed{\mathrm{A}}$.