The group $PSL(2,\mathbb{Z})=\mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/3\mathbb{Z}$ is not residually nilpotent. Indeed, if it was residually nilpotent, then it would be residually (finite $p$-)group since every finitely generated nilpotent group is residually finite (Malcev) and every finite nilpotent group is a direct product of finite $p$-groups. But in every finite $p$-group either the image of the generator $a$ of order 2 or the image of generator $b$ of order $3$ of $PSL(2,\mathbb{Z})$ is equal to 1 (since either 2 or 3 is co-prime with $p$). Hence the image of the commutator $[a,b]$ is 1. Thus one cannot separate $[a,b]$ from 1 in any nilpotent homomorphic image of $PSL(2,\mathbb{Z})$. 

On the other hand, for every $p$, any free product of two residually (finite $p$-) groups is residually (finite $p$-) by a result of Gruenberg (see, for example, http://www.ams.org/journals/bull/1969-75-02/S0002-9904-1969-12149-X/S0002-9904-1969-12149-X.pdf), and hence residually nilpotent.