Let $R$ be a commutative Ring with identity and $x\in R$ with $x=x^{2}$. Show that $\langle x \rangle$ is a Ring.
For me there are two possibilities coming to my mind, namely that x could be 1 or 0, however I think that this question demands an unambigous answer.
On
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.
EDIT: The post was edited so that $R$ is a commutative ring.
Note (as I failed to in the comments) that $\langle x \rangle$ being a ring need not make it a subring of $R$. Now as an ideal, $\langle x \rangle$ inherits every ring axiom from $R$ except for having identity. Now I claim that $x$ takes on the role of multiplicative identity in $\langle x \rangle$
Take $a \in \langle x \rangle$. Since $\langle x \rangle$ is an ideal, we can take $b \in R$ such that $a = bx$. Now since $x$ is idempotent and R is commutative we have: $$xa = ax = (bx)x = bx^2 = bx = a$$
Thus $x$ satisfies the definition of a multiplicative identity and $\langle x \rangle$ is a ring with operations inherited from $R$, additive identity $0 \in R$, and multiplicative identity $x$.