Here's a rather different argument. First, suppose that $R$ is commutative. Suppose $u+a$ is not a unit. Then it is contained in some maximal ideal $M\subset R$. Since $a$ is nilpotent, $a\in M$ (since $R/M$ is a field, and any nilpotent element of a field is $0$). Thus $u=(u+a)-a\in M$ as well. But $u$ is a unit, so it can't be in any maximal ideal, and this is a contradiction.
If you don't know that $R$ is commutative, let $S\subseteq R$ be the subring generated by $a$, $u$, and $u^{-1}$. Then $S$ is commutative: the only thing that isn't immediate is that $u^{-1}$ commutes with $a$, and this this can be proven as follows: $$u^{-1}a=u^{-1}auu^{-1}=u^{-1}uau^{-1}=au^{-1}.$$
The argument of the first paragraph now shows that $u+a$ is a unit in $S$, and hence also in $R$.
[The use of a maximal ideal (and hence the axiom of choice) in this answer can be avoided as follows. First, show that if $I$ is a proper ideal in a commutative ring and $a$ is nilpotent, then $I+(a)$ is still a proper ideal. Now take $I=(u+a)$ and let $M=I+(a)$, and argue as above. Note also that while this argument appears nonconstructive, it can be made constructive, and it is a fun and somewhat tricky exercise to chase through the proof and see how it can give an explicit inverse to $u+a$.]