If $N$ is the least normal subgroup of $A*B$ containing $A$, then $(A*B)/N \cong B$.
My Proof: Let $f:A \to B$ be the homomorphism given by $f(a) = e_B$. Note that $1_B:B \to B$ is also a homomorphism. Therefore, by the Universal Property of $A*B$, there exists a unique group homomorphism $\eta:A*B \to B$ where $\eta \circ i_A = f$ and $\eta \circ i_B = 1_B$ (where each $i_X$ is the inclusion map $i_X:X \to A*B$). Since $1_B = \eta \circ i_B$ is surjective, so is $\eta$. Additionally, note that $\ker(\eta) = \langle \langle A \rangle \rangle = N$ (I have no idea whether this is true or not). Therefore, by the First Isomorphism Theorem of Group Theory, there exists an isomorphism $(A * B)/\ker(\eta) \to B$. Therefore, $(A * B)/N \cong B$.
The part of the proof I'm having trouble with is finding a suitable homomorphism $f:A \to B$ so that $\ker(\eta) = N$. In particular, I am unsure how to show that $\ker(\eta) = N$ in any case. There is a question here asking about what the least normal subgroup is; however, it never gained any traction. Additionally, there is another question If $N$ is the normal subgroup of $A\ast B$ generated by $A$, then $(A\ast B)/N\cong B$, which is similar to mine.
Is my proof on the right track? How do I find a suitable homomorphism $f$ and how would I prove that $\ker(\eta) = N$? Thanks for any help.