Let $R$ be a ring and suppose $R$ is cyclic as a left $R$-module. That is there exists $\alpha \in R$ such that for any $r \in R$, there is some $r'$ such that $r'\alpha = r$. I want to show that $R$ has unity. That is $1 \in R$ such that $1r = r = r1$ for all $r \in R$. This is easy enough if $R$ is also assumed to be commutative, as shown below.
Suppose $R$ is commutative and cyclic as an $R$-module. Let $R = R\alpha$ for $\alpha \in R$. Then there exists $r_1 \in R$ such that $r_1\alpha = \alpha$. Let $r = r'\alpha \in R$. We have
$$r_1r = rr'\alpha = r' r \alpha = r'\alpha = r$$
Thus $r_1$ is the unity we desired.
I don't know how to prove this result if $R$ is not commutative, my gut tells me that there is probably some counter-example. Either a hint or a complete answer would be amazing.
You're right that it does not hold for noncommutative rings.
Let $R$ be the ring $M_2(\mathbb R)$, and let $e=\begin{bmatrix}1&0\\0&0\end{bmatrix}$. The set $Re=S$ is a rng without identity, and yet $Re=(Re)e$.