Does the unit generate the additive group in a unital ring with cyclic additive group?

EDIT My original answer was wrong. I've kept it below for completeness, but am writing a new, (hopefully) correct answer at the top.

The statement is true for all rings.

Let $R$ be a unital ring with cyclic additive group generated by $\alpha$. Then $R$ is commutative since $\alpha$ commutes with itself. Then $\alpha^2 = m\alpha$ for some integer $m$, which means that $(m - \alpha)\alpha = 0$. Now, $1 = k\alpha$ for some integer $k$, so $$ 0 = k\cdot 0 = k(m-\alpha)\alpha = (m-\alpha)k\alpha = (m - \alpha)\cdot 1= m - \alpha, $$ so $\alpha = m$, which means that $\alpha$ lies in the additive span of $1$, hence $1$ generates $(R, +, 0)$.

Original wrong answer below

The statement is false for finite and infinite rings.

For the finite case, take $R = \mathbb{Z}_6[X]/(2X - 1)$ and let $\alpha = X + (2X - 1) \in R$. The additive group of $R$ is generated by $\alpha$, but not by $1$.

For the infinite case, do the same thing with $R = \mathbb{Z}[X]/(2X - 1)$.

The answer's "yes" when the order of $1$ is finite. If $1$ has finite order strictly less than $|R|$, say $n$, then $n\cdot 1=0$ would imply that $n\cdot g=0$ for every element, but there is supposed to be an element of additive order strictly greater than (potentially even infinite) $1$'s order.

The answer is also yes when the order of $1$ is infinite (and hence $R$ is isomorphic to $\mathbb Z$), but the proof is different.

Let $g$ additively generate $R$. Then there exists some natural number $n$ such that $ng=1$, and another natural number $m$ such that $mg=g^2$.

Combining these two, $nmg=g$. But since $R$ is a free abelian group on $g$, this would mean $mn=1$, and the only possibilities are $n=m=1$ and $n=m=-1$, both of which imply $1$ is a generator.