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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.

Update 1. This is an answer to Andreas Thom's comment. Every free product $G=* G_i$ of Abelian groups is residually solvable. Indeed, $G$ has a homomorphism onto the direct product $A$ of $G_i$. The kernel $K$ certainly contains the derived subgroup (actually these two subgroups coincide but it does not matter here). The subgroup $K$ is free (hence the derived subgroup of $G$ is free). Indeed, if $K$ was not free, it would have a non-trivial intersection with one of the conjugates of $G_i$ by Kurosh's theorem, and hence a non-trivial image in $A$. Since $K$ is free, and $G/K$ is Abelian, $G$ is residually solvable. Moreover $G$ is residually (torsion-free nilpotent-by-Abelian).

Update 2. The question changed. Here is the answer to the new question. The free product of $F∗G$ where $F$ is free and $G$ is a free product of $\mathbb{Z}/2\mathbb{Z}$ is residually (finite 2-)group, hence residually nilpotent. For finitely many factors it follows from Gruenberg's result I mentioned above. For infinitely many factors take any word $w\ne 1$ in generators of the free product. Let $X$ be the finite set of generators that appear in $w$. Let $U$ be the finitely generated free factor of $F*G$ generated by $X$. Then $U$ is residually nilpotent by Gruenberg. Hence there exists a homomorphism $\phi$ onto a nilpotent group separating $w$ from 1. Since $U$ is a retract of $F*G$, the composition of the retraction and $\phi$ maps $F*G$ into a nilpotent group and separates $w$ from 1.

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.

Update 1. This is an answer to Andreas Thom's comment. Every free product $G=* G_i$ of Abelian groups is residually solvable. Indeed, $G$ has a homomorphism onto the direct product $A$ of $G_i$. The kernel $K$ certainly contains the derived subgroup (actually these two subgroups coincide but it does not matter here). The subgroup $K$ is free (hence the derived subgroup of $G$ is free). Indeed, if $K$ was not free, it would have a non-trivial intersection with one of the conjugates of $G_i$ by Kurosh's theorem, and hence a non-trivial image in $A$. Since $K$ is free, and $G/K$ is Abelian, $G$ is residually solvable. Moreover $G$ is residually (torsion-free nilpotent-by-Abelian).

Update 2. The question changed. Here is the answer to the new question. The free product of $F∗G$ where $F$ is free and $G$ is a free product of $\mathbb{Z}/2\mathbb{Z}$ is residually (finite 2-)group, hence residually nilpotent. For finitely many factors it follows from Gruenberg's result I mentioned above. For infinitely many factors take any word $w\ne 1$ in generators of the free product. Let $X$ be the finite set of generators that appear in $w$. Let $U$ be the finitely generated free factor of $F*G$ generated by $X$. Then $U$ is residually nilpotent by Gruenberg. Hence there exists a homomorphism $\phi$ from $U$ onto a nilpotent group separating $w$ from 1. Since $U$ is a retract of $F*G$, the composition of the retraction and $\phi$ maps $F*G$ into a nilpotent group and separates $w$ from 1.

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.

Update This is an answer to Andreas Thom's comment. Every free product $G=* G_i$ of Abelian groups is residually solvable. Indeed, $G$ has a homomorphism onto the direct product $A$ of $G_i$. The kernel $K$ certainly contains the derived subgroup (actually these two subgroups coincide but it does not matter here). The subgroup $K$ is free (hence the derived subgroup of $G$ is free). Indeed, if $K$ was not free, it would have a non-trivial intersection with one of the conjugates of $G_i$ by Kurosh's theorem, and hence a non-trivial image in $A$. Since $K$ is free, and $G/K$ is Abelian, $G$ is residually solvable. Moreover $G$ is residually (torsion-free nilpotent-by-Abelian).

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.

Update 1. This is an answer to Andreas Thom's comment. Every free product $G=* G_i$ of Abelian groups is residually solvable. Indeed, $G$ has a homomorphism onto the direct product $A$ of $G_i$. The kernel $K$ certainly contains the derived subgroup (actually these two subgroups coincide but it does not matter here). The subgroup $K$ is free (hence the derived subgroup of $G$ is free). Indeed, if $K$ was not free, it would have a non-trivial intersection with one of the conjugates of $G_i$ by Kurosh's theorem, and hence a non-trivial image in $A$. Since $K$ is free, and $G/K$ is Abelian, $G$ is residually solvable. Moreover $G$ is residually (torsion-free nilpotent-by-Abelian).

Update 2. The question changed. Here is the answer to the new question. The free product of $F∗G$ where $F$ is free and $G$ is a free product of $\mathbb{Z}/2\mathbb{Z}$ is residually (finite 2-)group, hence residually nilpotent. For finitely many factors it follows from Gruenberg's result I mentioned above. For infinitely many factors take any word $w\ne 1$ in generators of the free product. Let $X$ be the finite set of generators that appear in $w$. Let $U$ be the finitely generated free factor of $F*G$ generated by $X$. Then $U$ is residually nilpotent by Gruenberg. Hence there exists a homomorphism $\phi$ onto a nilpotent group separating $w$ from 1. Since $U$ is a retract of $F*G$, the composition of the retraction and $\phi$ maps $F*G$ into a nilpotent group and separates $w$ from 1.

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.

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.

Update This is an answer to Andreas Thom's comment. Every free product $G=* G_i$ of Abelian groups is residually solvable. Indeed, $G$ has a homomorphism onto the direct product $A$ of $G_i$. The kernel $K$ certainly contains the derived subgroup (actually these two subgroups coincide but it does not matter here). The subgroup $K$ is free (hence the derived subgroup of $G$ is free). Indeed, if $K$ was not free, it would have a non-trivial intersection with one of the conjugates of $G_i$ by Kurosh's theorem, and hence a non-trivial image in $A$. Since $K$ is free, and $G/K$ is Abelian, $G$ is residually solvable. Moreover $G$ is residually (torsion-free nilpotent-by-Abelian).