Yes, that's the idea. Let $A\subseteq G$ be measurable with respect to the Borel-algebra on $G$. Then $A\subseteq\mathbb{R}$ is measurable with respect to the Borel-algebra on $\mathbb{R}$ (Why?). This measure is inner regular, so $\lambda(A)=\sup\{\lambda(K)\colon K\subseteq A\text{compact}\}$. Now all the $K$ the supremum is taken over are subsets of $A$, hence subsets of $G$.
Now note that this is the same as saying that $A$ is an inner regular set in $G$, because a) a compact subset of $G$ is the same as a compact subset of $\mathbb{R}$ that is contained in $G$ and b) the restriction of…