 Solution

Fix the first tape of the group multiplier automaton $M_x$ to be $u=a_1\cdots a_m$. Construct its possible runs by dynamic programming, one column at a time. A record consists of a state and a flag indicating whether the output tape has ended. At column $i\le m$, the first symbol is $a_i$ and the second symbol may be any letter of the alphabet, or $\#$; after the second tape first uses $\#$ it must continue to use $\#$. At later columns the first symbol is $\#$, and an additional column must have a genuine letter on the second tape. Never add a $(\#,\#)$ column.

Keep one predecessor record and its output symbol for each reachable state-and-flag pair in each layer. Two runs reaching the same record have exactly the same possible future completions, so this merging loses no accepting output. The merged dead state of a finite automaton can be discarded. At each layer $i\ge m$, test whether a reached state is accepting; the initial layer is included when $m=0$. Upon success, follow predecessor records backwards and omit padding symbols to recover $v$.

The automatic structure for a group supplies a representative in $L$ of $\overline{u}x$, so an accepted padded convolution of words exists and this search terminates. The output is the required representative:The search uses only the finite transition table; it does not presuppose a solution to the word problem for a group.