Let us suppose that $A=\{a,b,\ldots z\}$ is a minimal set of generators for a finite group $G$. Then, does the sequence $a,ab,abc,\ldots, abc\ldots z, abc\ldots za, abc\ldots zab,\ldots\}$ produce all the distinct elements of $G$. Note that the $z$ being in the end is just symbolic. It does not imply that there are just $26$ generators, rather it only implies that the group is finitely generated.
I think yes, but am not clear how to show that all the words in the sequence are actually distinct. What if two words get the same element of $G$? How about if $G$ is the symmetric group $S_n$ and $A$ is the set of transpositions $\{(12), (13),\ldots, (1 n)\}$. Can we ensure at least in this case? Thanks beforehand.
If $n=|A|$ and $t$ is the order of the ordered product of the elements of $A$ (what you write $abc\cdots z$ but which, as noted in the comments, is bad notation, since it suggests there are $26$ elements), then clearly the sequence has at most $tn$ different elements. So in the case of $S_n$ at the end of your question, we will not get all elements for $n\geq 4$ at least.