Restating the question as I understand it:

Let $G$ be a group, $S\subset G$ a finite set of generators, $H$ a cyclic subgroup of $G$. Suppose that there is a finite collection of elements $g_j\in G$ such that every product of finitely many elements of $S$ belongs to one of the cosets $g_jH$. Is it true that $H$ has finite index in $G$?

The answer is yes, even without assuming $H$ cyclic. The group $G$ acts on the set $G/H$ of all cosets of $H$ in the usual fashion. The smallest subset of $G/H$ that contains the element $H$ and is closed under the action of each element of $S$ is finite. Therefore each element of $S$ maps this set to itself bijectively. Therefore so does the inverse of each element of $S$. Therefore so does each element of $G$. But the action of $G$ on $G/H$ is transitive, so it follows that this finite set is the whole of $G/H$, and $H$ has finite index.

thesubgroup generated by $g$»? – Mariano Suárez-Alvarez♦ Apr 7 '11 at 17:53