In general, if $G$ and $G^{\prime}$ are simply-connected (not necessarily compact), then $\mathfrak{g}\cong\mathfrak{g}^{\prime}$ implies $G\cong G^{\prime}$. Without simply connetedness, this is in general not true: $\mathfrak{su}(2)\cong\mathfrak{so}(3)$ but $\mathrm{SO}(3)$ is not isomorphic to $\mathrm{SU}(2)$. Simply connectedness is just a sufficient condition though.

@WillieWong I am quite late to this question, but it happens that I am also looking for exactly the same paper, but neither the library of my current affiliation nor other well-known internet sources seem to have this publication. Do you still have a copy and mind sharing it with me?

@TaQ Why does (strong) convergence in $(H^{s},\Vert\cdot\Vert_{H^{s}})$ (for $s\geq 0$) implies convergence in $(L^{2},\Vert\cdot\Vert_{L^{2}})$? Maybe I am missing something...

@TaQ I don't think it is so easy. If you take a differential operator $D:C^{\infty}_{c}\to C^{\infty}_{c}$, then in general, there is no relation between its closures in $H^{s}$ and $L^{2}$ in the sense that neither of the domains $\mathcal{D}(\overline{D}^{L^{2}})$ and $\mathcal{D}(\overline{D}^{H^{s}})$ is contained in each other (since there is no relation between strong convergence in $L^{2}$ and $H^{s}$). Of course, $\mathcal{D}(\overline{D}^{L^{2}})\cap \mathcal{D}(\overline{D}^{H^{s}})$ is non-empty and both operators agree on their common domain, but that is all we can say in general.

@TaQ I mean closure in the standard functional analytic sense: If $A:D(A)\to H$ is a linear operator in a Hilbert space $H$, then $\overline{A}$ is the operator defined as follows: $x\in D(\overline{A})$ iff there exists a sequence $(x_n)_n$ in $D(A)$ converging to $x$ such that $(Ax_n)_n$ is convergent. In this case we set $\overline{A}x:=\lim_{n\to\infty} Ax_n$. In other words, $\overline{A}$ is the (unique) operator whose graph is the closure of the graph of $A$.