Units and Nilpotents
If $u=1$, then you could do it via the identity $$(1+a)(1-a+a^2-a^3+\cdots + (-1)^{n}a^n) = 1 + (-1)^{n}a^{n+1}$$ by selecting $n$ large enough.
If $uv=vu=1$, does $a$ commute with $v$? Is $va$ nilpotent?
Let $v$ be the inverse of $u$, and suppose $a^2=0$. Note that $$(u+a)\cdot v(1-va)=(1+va)(1-va)=1-v^2a^2=1-0=1.$$ See if you can generalize this.
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 $R$.
This argument may seem horribly nonconstructive, due to the use of a maximal ideal (and hence the axiom of choice) and proof by contradiction. However, it can be made to be constructive and gives an explicit inverse for $u+a$ in terms of an inverse for $u$ and and an $n$ such that $a^n=0$.
First, we observe that all that is actually required of the ideal $M$ is that it is a proper ideal which contains $u+a$ and all nilpotent elements of $R$. So we may replace $M$ with the ideal $(u+a)+N$ where $N$ is the nilradical of $R$, and use the fact that if $I=(u+a)$ is a proper ideal in a commutative ring then $I+N$ is still a proper ideal. This is because $R/(I+N)$ is the quotient of $R/I$ by the image of $N$, which is contained in the nilradical of $R/I$. If $I$ is a proper ideal, then $R/I$ is a nonzero ring, so its nilradical is a proper ideal, so $R/(I+N)$ is a nonzero ring and $I+N$ is a proper ideal.
Next, we recast this argument as a direct proof instead of a proof by contradiction. Letting $I=(u+a)$, we observe that $I+N$ is not a proper ideal since $u=(u+a)-a\in I+N$ and $u$ is a unit. That is, a nilpotent element (namely $a$) is a unit in the ring $R/I$, which means $R/I$ is the zero ring, which means $I=R$, which means $u+a$ is a unit.
Finally, we chase through the explicit equations witnessing the statements above. Letting $v=u^{-1}$, we know that $v((u+a)-a)=1$ so $$-va=1-v(u+a),$$ witnessing that $a$ is a unit mod $u+a$ (with inverse $-v$). But $a$ is nilpotent, so $a^n=0$ for some $n$, and thus $0$ is also a unit mod $u+a$. We see this explicitly by raising our previous equation to the $n$th power: $$0=(-v)^na^n=(1-v(u+a))^n=1-nv(u+a)+\binom{n}{2}v^2(u+a)^2+\dots+(-v)^n(u+a)^n,$$ where every term after the first on the right-hand side is divisible by $u+a$. Factoring out this $u+a$, we find that $$1=(u+a)\left(nv-\binom{n}{2}v^2(u+a)+\dots-(-v)^n(u+a)^{n-1}\right)$$ and so $$-\sum_{k=1}^n \binom{n}{k}(-v)^k(u+a)^{k-1}= nv-\binom{n}{2}v^2(u+a)+\dots-(-v)^n(u+a)^{n-1}$$ is an inverse of $u+a$.
The fact that this complicated formula is hidden in the one-paragraph conceptual argument given at the start of this answer is a nice example of how powerful and convenient the abstract machinery of ring theory can be.