There is a general argument that is slightly more elementary than what you wrote. By standard properties of Weil restrictions, we have $R(L) = \prod_{\sigma} \mathbb{G}_m(L)$, where the product is taken over the different $K$-linear embeddings of $L$ into some algebraic closure $\overline{K}$ and where $G$ acts on the factors in a natural way. In other words, we have that $R(L) = \mathbb{Z}[G] \otimes \mathbb{G}_m(L)$, as $G$-modules. By an easy exercise, we have that $$\mathbb{Z}[G] \otimes \mathbb{G}_m(L) = \operatorname{Ind}^G_{\{1\}}(\mathbb{G}_m(L)),$$ giving $R(L)=\operatorname{Ind}^G_{\{1\}}(\mathbb{G}_m(L))$. (Indeed, take the isomorphism $$ f:\mathbb{Z}[G] \otimes \mathbb{G}_m(L) \rightarrow \mathbb{Z}[G] \otimes (\mathbb{G}_m(L))_0 \stackrel{\operatorname{def}}{=} \operatorname{Ind}^G_{\{1\}}(\mathbb{G}_m(L)), $$ where $\mathbb{G}_m(L)_0$ denotes the underlying abelian group of $\mathbb{G}_m(L)$, satisfying $f(g \otimes x) = g \otimes g^{-1}x$.)
In conclusion, $R(L)=\operatorname{Ind}^G_{\{1\}}(\mathbb{G}_m(L))$. Finally, by Shapiro's lemma, the $\operatorname{H}^1$ (and in fact, all higher cohomology) of this vanishes.