There is an exact sequence $\operatorname{SL}(2,R)_{ab}\rightarrow \operatorname{GL}(2,R)_{ab} \xrightarrow{\ \det\ }R^*\rightarrow 1 $$\operatorname{SL}(2,R)_\text{ab}\rightarrow \operatorname{GL}(2,R)_\text{ab} \xrightarrow{\ \det\ }R^*\rightarrow 1 $. Now, if $R$ is not a ring of imaginary quadratic integers, $\operatorname{SL}(2,R)_{ab}\otimes \mathbb{Q} $$\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 $Le problème des groupes de congruence pour $\operatorname{SL}_2 $ by J.-P. Serre, Ann. Math. 92, no. 3, 489-527489–527 (1970). Therefore $\ \det\ $$\det$ induces an isomorphism $\operatorname{GL}(2,R)_{ab}\otimes \mathbb{Q}\rightarrow R^*\otimes \mathbb{Q}\ $$\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.
