I saw this problem in my class notes as an exercise and I’ve been stuck for a bit now. My intuition is that in general the statement is false. If the ring has unity, then it seems that the statement would be true, since for any nonzero element <1>=R.
So I’ve been trying to look at rings upper triangular matrices with entries from <0,2> where addition and multiplication is mod4, so that the ring doesn’t have unity and a non commutative ring, but I’m unsure what $_RR$ would look like here. My thinking is that since multiplication on the left by any element gives back the zero matrix, we'd just get 0?
Am I on the right track with thinking, or am I way off?
For $_RR$ to be cyclic, it'd have to be of the form $\mathbb Z r+Rr$ for some element $r\in R$. (Of course, this just becomes $Rr$ if $R$ has identity.)
With that in mind, we could choose $R$ to be such that $R^2=\{0\}$, and try to ensure that $R\neq \mathbb Zr$ for any $r$. That way if you supposed that $R=\mathbb Zr+Rr$ for some $r$, you could see this is a contradiction, because $Rr=\{0\}$ and $R\neq \mathbb Zr+\{0\}$.
Now, your example works for these reasons, but you can also get away with a somewhat simpler description: how about $R=(x)/(x^2)$ where $(x)$ and $(x^2)$ are the obvious ideals in $\mathbb Q[x]$?
OTOH, the matrix ring has you described it has only 8 elements, all of additive order $2$, so it's a good one to give also.