What are the units in $R[X,X^{1}]$, where $R$ is a commutative ring with $1$? I know that the question for polynomial rings is a standard textbook exercise. However, I couldn't find a reference for Laurent polynomials, since most people only seem to consider coefficients in an integral domain.

You can find a more general result in the paper [1], which determines the units and nilpotents in arbitrary group rings $\rm R[G]$ where $\rm G$ is a uniqueproduct group  which includes ordered groups. As the author remarks, his note was prompted by an earlier paper [2] which explicitly treats the Laurent case. 1 Erhard Neher. Invertible and Nilpotent Elements in the Group Algebra of a Unique Product Group 2 Ottmar Loos. Remarks on Holger P. Petersson's "Idempotent 2by2 matrices" http://homepage.uibk.ac.at/~c70202/jordan/archive/remarks/remarks.pdf 


Thanks for your comments! A colleague just pointed me to an article by Karpilovsky ('On finite generation of unit groups of commutative group rings'). Translated into our setup, this gives the following result: Take $r \in R^\times$, $a_i \in R$ nilpotent ($i \in \mathbb{Z}$), $k\ge 0$,$e_1,\dots,e_k \in R$ orthogonal idempotents that sum up to $1$, and fix $i_1,\dots,i_k \in \mathbb{Z}$. Then the element $$ r (1+\sum_i a_i X^i )(e_1X^{i_1} + \dots + e_kX^{i_k} ) $$ is a unit in $R[X,X^{1}]$ and all units arise in this way. In particular, all units are of the form $rX^j$ with $r \in R^\times$iff $R$ is reduced and connected, as Boyarsky pointed out. I was hoping for a nice condition on the coefficients for the polynomial (as in the case $R[X]$). Maybe someone still sees how to simplify this statement or elegantly prove it in this setup? 


Although the proof is only sketched in Exercise 3.7 of The Kbook: An introduction to algebraic Ktheory (Graduate Studies in Mathematics 145) by Charles A. Weibel I found this reference very helpful (see how the direct product decomposition of the unit group is made explicit there). 

