Let $R$ be an integral domain. I am looking for a general way to describe the units in $$ R[x,x^{-1}] := R[x,y]/(xy-1).$$
Clearly $$\{rx^n \mid n \in \mathbb{Z}, r \in R^\times \} \subseteq R[x,x^{-1}]^\times$$
is a subgroup, but how do I know whether it's all? I was trying to argue with degrees or to write down multiplication of general elements with Cauchy sums, but it all becomes pretty messy, having also negative exponents.
When $R$ is not integral it's not necessarily true that this is all there is : look at $R= \mathbb Z/4$ and $(2x+1)^2 = 4x^2+4x+1 = 1$ : even in $R[x]$ there can be other units.
If $R$ is an integral domain, those are indeed the only ones : take $fg = 1$, let $k$ (resp. $j$) be the highest index for which $f_k$ (resp. $g_j$) is nonzero, $k',j'$ similarly but with lowest.
Then $f_kg_j x^{k+j}$ is the highest degree monomial of $fg$, and $f_kg_j \neq 0$ by integrality. It follows that $k+j = 0$ and $f_kg_j = 1$.
Similarly, $f_{k'}g_{j'}x^{k'+j'}$ is the lowest degree monomial of $fg$ and $f_{k'}g_{j'} \neq 0$ by integrality. It follows that $k'+j'= 0$.
But $k'\leq k, j'\leq j$ by definition, therefore since both sums are $0$ we must have $k=k', j=j'$ and in particular, $f$ has only one nonzero monomial (so $f= rx^n$ for some $r,n$), similarly for $g$, and the rest follows.
Another way to view your ring is as $R[x]$ localised at the multiplicative set $S = \{r\cdot x^k\mid r\in R^\times, k\in \Bbb N\}$. Really, it's just localised at the set $\{1, x, x^2, \ldots\}$, but including the units of $R$ allows us to skip a step.
I claim that this $S$ is saturated, which is to say if $f, g\in R[x]$ are such that $fg\in S$, then $f, g\in S$. This is not difficult to see:
which is to say $f, g\in S$.
Why does this matter? Because if you localise by a saturated set $S$, then that saturated set contains all elements that become invertible in the localisation.
Let $\frac fs\in S^{-1}R[x]$ be invertible. Then there is a $\frac gt\in S^{-1}R[x]$ such that $\frac fs\cdot \frac gt = \frac11$. By definition of localisation, this means that $$ fg\cdot 1 = st\cdot 1 $$ (The absence of zero divisors in $R[x]$ means we can simplify the definition of equality in a localisation to this.) Since $st\in S$ because $S$ is multiplicatively closed, we also have $fg\in S$. However, this implies that $f\in S$. So $\frac fs$ is a fraction where both he numerator and the denominator are in $S$, and because of the general form of elements in $S$, this means that $\frac fs$ is either equal to a fraction of the form $\frac{1}{rx^k}$ or to a fraction of the form $\frac{rx^k}1$ where $r\in R^\times, k\in \Bbb N$.
