This is a re-statement, of sorts, of http://mathoverflow.net/questions/71389, so far unanswered.
Let $G$ be a Polish group, $d_L$ a compatible left-invariant metric on $G$. This metric is usually not complete, so let $\hat G$ be the completion of $G$ with respect to $d_L$. If $(g_i)$ and $(h_i)$ are Cauchy sequences in $(G,d_L)$ then so is $(g_i h_i)$, endowing $\hat G$ with a semi-group structure.
Since any two left-invariant compatible metrics on $G$ are uniformly equivalent, none of this depnds on the precise choice of $d_L$.
Question: say $a,b \in \hat G$. Are there always $c,d \in \hat G$ such that $ca = db$? (No idea why this should be true, but then what is a counter-example?)
Motivation: $G$ can always be viewed as the automorphism group of some complete separable approximately ultra-homogeneous metric structure $M$, and $G$ is a closed subgroup of $S_\infty$ if and only if $M$ can be taken to be a countable ultra-homogeneous discrete structure (what logicians usually understand by "structure"). Then $\hat G$ is the semi-group of embeddings of $M$ in itself. Now the question becomes very close (and in the discrete case, possibly equivalent) to the one cited above: can any two copies of $M$ be amalgamated over a common copy of $M$, with the result embeddable in $M$?