 Solution

No such ring exists: every finite integral domain is a field. Here is the proof. For $a\ne 0$, the map $L_a:R\to R$ defined by $L_a(x)=ax$ is injective, since $ax=ay$ gives $a(x-y)=0$ and therefore $x=y$. An injective self-map of a finite set is surjective, so some $x$ satisfies $ax=1$. Since an integral domain is commutative, $xa=1$ as well. Thus every nonzero element has a multiplicative inverse, making $R$ a field.

The conclusion still holds if an identity was not initially required. Surjectivity first gives an element $e$ with $ae=a$. Then $a(ex-x)=0$ implies $ex=x$ for every $x$, and commutativity makes $e$ a two-sided identity. Applying the preceding argument with $e$ supplies all inverses.