The free product $\mathbb Z_2$ and $\mathbb Z_3$ (i.e. PSL(2, $\mathbb Z$) is Gromov-hyperbolic (as every virtually free group) and non-virtually cyclic. Therefore by a result of Olshanskii, "SQ-universality of hyperbolic groups". (Russian)  Mat. Sb.  186  (1995),  no. 8, 119--132;  translation in  Sb. Math.  186  (1995),  no. 8, 1199–1211, it is SQ-universal, that is every countable group embeds into a factor group of PSL(2, $\mathbb Z$). In "most" of these groups (by construction) $ab$ will have infinite order. Thus, in particular, there are uncountably many groups of the type you want.  

Update: An explicit example would be this. Take $G=PSL(2,\mathbb Z)$, and any word $w(a,b)$ satisfying very small cancelation (that it no subword of length, say, $\frac{1}{10000}|w|$ occurs twice in $w$ (considered as a cyclic word). Then consider the group $G/\langle\langle w\rangle\rangle$. It is what you want. Geometrically, you just kill the large loop in the standard $K(\pi,1)$ for $PSL(2,\mathbb{Z})$ of course. 

Another example, as far as I remember, is the R. Thompson group $V$ (it is generated by an element $a$ of order 2 and an element $b$ of order 3 such that $ab$ has infinite order (Mason?). It should be written in the Cannon-Floyd-Parry's survey on Thompson groups, but I do not have it with me.