There is an exact sequence $\operatorname{SL}(2,R)_\text{ab}\rightarrow \operatorname{GL}(2,R)_\text{ab} \xrightarrow{\ \det\ }R^*\rightarrow 1  $. Now, if $R$ is not  a ring of imaginary quadratic integers, $\operatorname{SL}(2,R)_\text{ab}\otimes \mathbb{Q} $ is zero: this follows from the Corollary to Theorem 3 in [*Le problème des groupes de congruence pour $\operatorname{SL}_2 $*](https://doi.org/10.2307/1970630) by J.-P. Serre, Ann. Math.
92, no. 3, 489–527 (1970). Therefore $\det$ induces an isomorphism $\operatorname{GL}(2,R)_\text{ab}\otimes \mathbb{Q}\rightarrow R^*\otimes \mathbb{Q}$ in that case. You'll find how to treat the imaginary quadratic case in the same paper.