So I'm having some issues on fully believing my response to the following:
Let $R$ be a commutative ring. Is $xR$ always an ideal? Prove or show a counterexample.
So at first glance I believe this is true because if we have a right ideal, then by commutativity it follows that it is also a left ideal and thus a two sided ideal. But I also have this thought that unity has some importance for ideals but I have not found a counterexample to this statement with this key property.
Yes, it is always an ideal. As for the importance of unity, two things come to my mind that may have something to do with this.
Assuming all rings are required to be unital, a subring would have to contain the unity, but not an ideal. Maybe you're confusing these two definitions with each other?
Also, there's this property (theorem, lemma) that if an ideal contains the unity, then it's the entire ring.
The subset $xR=\{xr:r\in R\}$ is certainly an ideal of $R$:
Thus $xR$ is a right ideal, which is automatically also a left ideal in case $R$ is commutative.
What can fail when $R$ is not supposed to have an identity, is that $x\in xR$. For instance in the ring (without identity) $R=2\mathbb{Z}$, if $x=4$, then $x\notin xR$, because $4(2\mathbb{Z})=8\mathbb{Z}$.
In this case, if you want the least ideal containing $x$, you have to consider $$ xR+\mathbb{Z}x=\{xr+nx:r\in R, n\in\mathbb{Z}\}. $$ When $R$ has an identity, $xR+\mathbb{Z}x=xR$.
If you mean a proper ideal, then $x$ must be non-invertible, but if you just say an ideal, then $xR$ is always an ideal.