Abelianization of $\mathbb{Z}_2*\mathbb{Z}_3$ Intuitively it has to be
$$\text{Ab}(\mathbb{Z}_2*\mathbb{Z}_3)=\mathbb{Z}_2\times\mathbb{Z}_3$$
here is my approach on how to prove it
$$\mathbb{Z}_2=P(a\mid a^2),\mathbb{Z}_3=P(b\mid b^3)\Rightarrow \mathbb{Z}_2*\mathbb{Z}_3=P(a,b\mid a^2,b^3)$$
then
$$\text{Ab}(\mathbb{Z}_2*\mathbb{Z}_3)=P(a,b\mid 2a,3b)=\mathbb{Z}\langle a,b\rangle/\langle 2a,3b\rangle=\mathbb{Z}\langle a\rangle/\langle 2a\rangle\times\mathbb{Z}\langle b\rangle/\langle 3a\rangle$$
which gives the wanted result, although I don't why that last equality holds.
 A: The last step of your proof follows from the fact that for groups $G_1$ and $G_2$ and normal subgroups $N_1 \subseteq G_1$ and $N_2 \subseteq G_2$ we have that $N_1 \times N_2 \subseteq G_1 \times G_2$ is a normal subgroup with $(G_1 \times G_2)/(N_1 \times N_2) \cong (G_1 / N_1) \times (G_2 / N_2)$.
To see the isomorphism notice that the group epimorphism
$$
\tilde{\varphi} \colon G_1 \times G_2 \to (G_1/N_1) \times (G_2/N_2),
\quad
(g_1,g_2) \mapsto ([g_1],[g_2])
$$
has kernel $N_1 \times N_2$ (which is therefore normal), and thus factors through a group isomorphism
$$
\varphi \colon
(G_1 \times G_2)/(N_1 \times N_2) \to (G_1/N_1) \times (G_2/N_2)
 \quad
[(g_1, g_2)] \to ([g_1],[g_2])
$$

I think that $\operatorname{Ab}(\mathbb{Z}/2 * \mathbb{Z}/3) \cong \mathbb{Z}/2 \oplus \mathbb{Z}/3$ can also nicely be seen by using the universal properties of the appearing objects, leading to a more general result:
If $A_1$ and $A_2$ are abelian groups then $\operatorname{Ab}(A_1 * A_2)$ together with the two maps $i_1 \to \operatorname{Ab}(A_1 * A_2)$, $a_1 \mapsto [a_1]$ and $i_2 \colon A_2 \to \operatorname{Ab}(A_1 * A_2)$, $a_2 \mapsto [a_2]$ satisfies the universal property of the direct sum, i.e. the coproduct in the category of abelian groups:
Given any abelian group $B$ and group homomorphisms $f_1 \colon A_1 \to B$ and $f_2 \colon A_2 \to B$ there exists, by the universal property of the free product, a unique group homorphism $\tilde{g} \colon A_1 * A_2 \to B$ with $\tilde{g} \circ j_1 = f_1$ and $\tilde{g} \circ j_2 = f_2$, where $j_1 \colon A_1 \to A_1 * A_2$ and $j_2 \colon A_2 \to A_1 * A_2$ denote the inclusions into the free product.
Because $B$ is abelian $\tilde{g}$ factors through a unique group homomorphism $g \colon \operatorname{Ab}(A_1 * A_2) \to B$ by the universal property of the abelization. $g$ is the unique group homomorphism $h \colon \operatorname{Ab}(A_1 * A_2) \to B$ with $h \circ i_1 = f_1$ and $h \circ i_2 = f_2$.
This shows that $\operatorname{Ab}(A_1 * A_2)$ together with $i_1$ and $i_2$ satisfies the universal property of the direct sum. As the direct sum is unique up to unique isomorphism there exists a unique isomorphism $\varphi \colon \operatorname{Ab}(A_1 * A_2) \to A_1 \oplus A_2$ with $\varphi \circ i_1 = \iota_1$ and $\varphi \circ i_2 = \iota_2$, where $\iota_1 \colon A_1 \to A_1 \oplus A_2$, $a_1 \mapsto (a_1,0)$ and $\iota_2 \colon A_2 \to A_1 \oplus A_2$, $a_2 \mapsto (0,a_2)$. More explicitely, $\varphi([a_1 a_2]) = (a_1, a_2)$ for all $a_1 \in A_1$, $a_2 \in A_2$.

PS: This tells us that the functor $\operatorname{Ab} \colon \mathbf{Grp} \to \mathbf{Ab}$ is somehow compatible with coproducts, which can probably be formulated in a fancier and more category theoretic way.
A: This is most easily proved by abstract nonsense.
There's a forgetful functor $U : \mathbf{Ab} \rightarrow \mathbf{Grp}$. It has a left adjoint $F : \mathbf{Grp} \rightarrow \mathbf{Ab},$ the abelianization functor. Since left-adjoints commute with colimits, we have: $$F(G \sqcup H) = F(G) \sqcup F(H)$$
But an abelian group is just a $\mathbb{Z}$-module. Hence $\mathbf{Ab}$ has biproducts. So: 
$$F(G \sqcup H) = F(G) \times F(H)$$
