Is it possible that $(ab)^{-1}$ is defined although $a^{-1},b^{-1}$ are not?
Let $R = L(\ell^2)$ the linear continuous operators on $\ell^2$. Define $a,b \colon \ell^2 \to \ell^2$ by $$ a(x_1, x_2, \ldots) = (x_2, x_3, \ldots) $$ and $$ b(x_1, x_2, \ldots) =(0, x_1, x_2, \ldots) $$ Then $ab = 1$ is the identity, hence invertible, but neither $a$ nor $b$ are units as $a$ is not one to one, where $b$ is not onto.
It is not possible in commutative rings. In general, if $ab$ is a unit then there exists, $v\in R$ such that $(ab)v=1$. Thus $a(bv)=1$. So, $a$ is a right unit. S Similarly it can be shown that $b$ is a left unit.