Semigroup product of the left-invariant completion of a Polish group (restatement of Question 71389)

Question

This is a re-statement, of sorts, of the question Is there a relational countable ultra-homogeneous structure whose countable substructures do not have the amalgamation property?, 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 semigroup structure.

Since any two left-invariant compatible metrics on $G$ are uniformly equivalent, none of this depends on the precise choice of $d_L$.

Question: Given $a,b \in \hat G$, 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 $M$?

Answer 1

In the end it was the original question which was answered first.

The answer to Is there a relational countable ultra-homogeneous structure whose countable substructures do not have the amalgamation property? by Ali Enayat shows that there exists a countable ultra-homogeneous structure $M$ with embeddings of $f_i\colon M \to M$, $i = 0,1$, which do not amalgamate inside $M$.

Taking $G = \textrm{Aut}(M)$, $G$ is a Polish group (and moreover homeomorphic to a closed subgroup of $S_\infty$), $f_i \in \hat G$, and there are no $g_i \in \hat G$ such that $g_0 f_0 = g_1 f_1$. This gives the desired counter-example.

(Thank you, Ali!)