Because $xx^3x=x$ for all $x$, the ring is von Neumann regular.
Exercise: $R$ also has no nonzero nilpotent elements.
Exercise: If there are no nonzero nilpotents, that implies all idempotents are central.
Exercise: the idempotents are partially ordered by the relation $e\leq f$ iff $ef=e$.
Exercise: every two idempotents have a common upper bound . Hint: $e+f-ef$
Exercise: By finiteness, there is a unique maximal idempotent with respect to that order, and it is the identity of the ring (Hint: Let $e$ be a maximal idempotent. For arbitrary $a$, there exists $x$ such that $axa=a$. Notice that $ax$ and $xa$ are idempotents and look first at $ea=eaxa$.)
Exercise: a finite von Neumann regular ring is semisimple (having an identity speeds this along, somewhat.)
Since the ring has no nonzero nilpotents, it is a finite product of division rings. (Using Artin-Wedderburn.)
Since finite division rings are fields, it is a finite product of finite fields. (Wedderburn's little theorem.)
In each such field, since every element is a root of $x^5-x$, each of these fields can have no more than $5$ elements.
You can verify that $F_5, F_3, F_2$ all work, but $F_4$ does not.
So, now you should be able to see completely what such a ring looks like. I leave it to you to put into words.
Alternatively, you can apply a sledgehammer due to Jacobson to say very early on that $R$ is commutative, and find an identity the same way. I tried to avoid that.