From the Lie algebra inclusion $\mathfrak{g}\rightarrow \Gamma(T_G)$ of right invariant vector fields we get a map $U(\mathfrak{g})\rightarrow\Gamma(\mathcal{D}_G).$ This induces a map $U(\mathfrak{g})\otimes\mathcal{O}_G\rightarrow\mathcal{D}_G$ of $\mathcal{O}_G$-modules. We have a structure of $G$-equivariant sheaf on both sides, with respect to the right translation action of $G$ on itself. (The $G$-equivariant structure on the left hand side comes from tensoring up the $G$-equivariant structure on $\mathcal{O}_G,$ i.e., the $G$ action on the $U(\mathfrak{g})$ part is trivial.) Our map is $G$-equivariant, and if we know it is an isomorphism, we will get an isomorphism $\Gamma(U(\mathfrak{g})\otimes\mathcal{O}_G)^G\rightarrow\Gamma(\mathcal{D}_G)^G,$ or $U(g)\rightarrow\Gamma(\mathcal{D}_G)^G$, as desired.

Now let us show that $U(\mathfrak{g})\otimes\mathcal{O}_G\rightarrow\mathcal{D}_G$ is an isomorphism. We have a canonical filtration $F_i\mathcal{D}_G$ on $\mathcal{D}_G$ by order of differential operators. On the left hand side, the PBW filtration $F_iU(\mathfrak{g})$ induces a filtration $F_iU(\mathfrak{g})\otimes\mathcal{O}_G$.

As our map sends $F_iU(\mathfrak{g})\otimes\mathcal{O}_G$ to $F_i\mathcal{D}_G$ and both filtrations are exhaustive, it suffices to check that the map on associated gradeds is an isomorphism. The associated graded of the left hand side is $\operatorname{Sym}^{\bullet}\mathfrak{g}\otimes\mathcal{O}_G.$

For the right hand side, we know that $F_i\mathcal{D}_G/F_{i-1}\mathcal{D}_G$ can be canonically identified with $\operatorname{Sym}^iT_G.$ The inclusion $\mathfrak{g}\rightarrow \Gamma(T_G)$ gives rise to a map $\mathfrak{g}\otimes\mathcal{O}_G\rightarrow T_G$ which can be seen to be an isomorphism. This identifies the associated graded of the right hand side with $\operatorname{Sym}^{\bullet}\mathfrak{g}\otimes\mathcal{O}_G,$ and the map on associated gradeds becomes the identity.