The answer to this question can be found in Lubos Motl's answer to [this question of mine on Physics.SE](https://physics.stackexchange.com/q/65092/23119). 

The key here are the weight lattices bosonic representations $\Gamma$ of these gauge groups.

As I understand it, the weight lattice of $E(8)$ is $\Gamma^8$, whereas the weight lattice of $\frac{\operatorname{Spin}\left(32\right)}{\mathbb{Z}_2}$^ is $\Gamma^{16}$. The first fact means that the weight lattice of $E(8)\times E(8)$ is $\Gamma^{8}\oplus\Gamma^8$,    

Now, an identity, that $\Gamma^{8}\oplus\Gamma^8\oplus\Gamma^{1,1}=\Gamma^{16}\oplus\Gamma^{1,1} $ , which actually allows this T-Duality. Now, this means that it is *this very identity* which allows the identity mentioned in the original post. 

So, the answer to your question is "**Yes**", there *is* a group-theoretical fact, and that is that   $ \Gamma^{8}\oplus\Gamma^8\oplus\Gamma^{1,1}= \Gamma^{16}\oplus\Gamma^{1,1} $.