Mark Sapir has already given you a definitive answer, but I think it is interesting to note that his argument generalises to show something stronger. A finitely-generated non-trivial free product of groups is residually nilpotent only if it is $p^{\prime}$-torsion-free, for some prime $p$. For (Here, $p^{\prime}$ denotes the set of primes not equal to $p$ so, in this case, we are allowed only $p$-torsion.) For, in a finitely-generated residually nilpotent group, elements of finite coprime order commute. And, if $G = A\ast B$ is such a free product (with $A\neq 1\neq B$), and $G$ is not $p^{\prime}$-torsion-free, for any prime $p$, there are distinct primes $p$ and $q$, and elements $u$, of order $p$, and $v$, of order $q$ in $G$. These may be taken (up to conjugation) to be members of $A\cup B$. Now, if $u$ and $v$ belong to different free factors ($A$ or $B$) then they cannot commute, so it must be that $u$ and $v$ belong to the same factor, say $A$. But then, taking a non-trivial element $b$ in $B$ and forming the conjugate $w = b^{-1}vb$, we get an element of order $q$ which again does not commute with $u$. (The commutator $[u,w] = u^{-1}b^{-1}v^{-1}bub^{-1}vb\neq 1$, by the normal form theorem for free products.) This again contradicts the fact that elements of coprime order commute.
