A divisor of a unit is a unit? Is it true that if $ab=u$ where $u\in U(R)$ is a unit of the noncommutative ring $R$, then $a,b\in U(R)$? If $R$ is commutative, then this can be seen by
$$a(bu^{-1})=uu^{-1}=1=(bu^{-1})a,$$
but when $R$ is not commutative I am not seeing an easy identity like the above. Is there a counterexample?
 A: No. There exists a ring with $a,b$ such that $ab=1$ and $ba\neq 1$. Even though $1$ is a unit, neither $a$ nor $b$ are units.
Looks like I had trouble finding this before when I was writing this solution:

Examples of such rings with $a$ and $b$ like that are scattered throughout the site, but a little hard to search for. I found one here at this related question, although you don't really need to say "bounded linear operators on $\ell^2$," you can just say "linear transformations from $V\to V$ where $V$ is a vector space with countably infinite dimension." For a fixed basis, the "right shift" $a$ and the "left shift" $b$ on the basis elements create linear transformations such that $ba=1$ and $ab\neq 0,1$.

That is, given a basis $\{v_i\mid i\in \Bbb N\}$, $a$ maps $v_i\mapsto v_{i+1}$, and the other map $b$ maps $v_i\mapsto v_{i-1}$ when $i\geq 1$, and it just maps $v_0$ to zero.
A possible fix
But if both $ab$ and $ba$ are units, then $a$ and $b$ are units. Suppose $abx=1$ and $yba=1$. Then $a^{-1}=bx=yb$. Finally, $a^{-1}ab=b$ is a unit because it is a product of two units.
A: Following up on @rschwieb's answer:
$R = L(\ell^2)$ (ring of linear transformations), $a$ = left shift $(x_1,x_2, \dots) \mapsto (x_2,x_3, \dots)$, $b$ = right shift  $(x_1,x_2, \dots) \mapsto (0, x_1,x_2, \dots)$.
