I'm going to use $e$ instead of $x$ because it is more common.
The general situation is that $eRe$ is a ring with identity $e$, which you can easily check. It is not necessarily an ideal, however.
If you don't require that a subring has identity, then this is all trivial since ideals are "subrings" in that case.
In general it is false that $\langle e \rangle$ is a ring with identity. For example, in the ring of linear transformations of a countably infinite dimensional vector space, you can pick an idempotent element given by a transformation which projects onto a single $1$-d subspace. The ideal generated by $e$ is the only nontrivial ideal of the endomorphism ring, and it does not have an identity.
If you were secretly assuming $R$ is commutative (or, more interestingly, when $e$ is assumed to be central), then yes, $eRe=\langle e \rangle = eR$ and it is, in fact, a subring with identity $e$.
There, you have your unambiguous answer. Now you can go ahead and verify axioms.