Has the Roelcke completion of a topological group any reasonable algebraic structure?

Question

It is well-known that each topological group $G$ carries (at least) four natural uniformities:

the left uniformity $\mathcal L$, generated by the base $\{\{(x,y)\in G\times G:y\in xU\}:U\in\mathcal \tau_e\}$;

the right uniformity $\mathcal R$, generated by the base $\{\{(x,y)\in G\times G:y\in Ux\}:U\in\mathcal \tau_e\}$;

the two-sided uniformity $\mathcal L\vee\mathcal R$, generated by the base $\{\{(x,y)\in G\times G:y\in xU\cap Ux\}:U\in\mathcal \tau_e\}$;

the Roelcke uniformity $\mathcal L\wedge\mathcal R$, generated by the base $\{\{(x,y)\in G\times G:y\in UxU\}:U\in\mathcal \tau_e\}$.

Here by $\tau_e$ we denote the family of all open neighborhoods of the unit $e$ in $G$.

It is well-known that the completion of $G$ by its two-sided uniformity $\mathcal L\vee\mathcal R$ carries a natural structure of a topological group, and the completion of $G$ by its left (or right) uniform structure carries a structure of a topological semigroup.

Question. Has the completion $\bar G$ of a topological group $G$ by its Roelcke uniformity any reasonable algebraic structure? For example, is $\bar G$ a semigroup with continuous right and left shifts by elements of the group $G$?

Remark. In Example 4.3 Uspenskii observed that for the homeomorphism group $G=H(2^\omega)$ of the Cantor cube $2^\omega$ the Roelcke completion of $G$ can be identifies with the semigroup of closed relations on $R\subset 2^\omega\times 2^\omega$ having both projections onto $2^\omega$ surjective. This semigroup is not semitopological and not inverse, so there is no hope that the Roelcke completion of any topological group carries the structure of a semitopological (inverse) semigroup. But at least some algebraic structure should be present?