Proving there exists a unique group structure such that there exists a homomorphism $A \to A/{\sim}$.

Question

Here is the context for this problem.

Let $A$ be a group, $A' \subset A$ a subgroup, and define $\sim^{l}$ and $\sim^r$, the left and right relations, respectively, on $A$ with respect to $A'$ by declaring $a \sim^{l} b$ if there exists $a' \in A'$ such that $b = a' a$ and $a \sim^{r} b$ if there exists $a'' \in A'$ such that $b = a a''$.

I proved that $A'$ is normal in $A$ if and only if $\sim^{l}$ and $\sim^{r}$ coincide. . If $A'$ is normal, then the left and right quotient sets are equal (though, as I understand, it is not the case that every element commutes; using more familar notation, if $H \leq G$ is a subgroup, then $gH = Hg$ for all $g \in G$, but it is not the case that given $h \in H$, $gh = hg$ for all $g \in G$.)

Finally, given a group $A$, let $\pi: A \to A/{\sim}$ be the canonical projection $a \mapsto \overline{a}$, where $\overline{a}$ is the coset containing $a$.

I am trying to prove the following theorem.

If $A'$ is a normal subgroup of $A$, the set $A/A'$ has a unique group structure for which the map $\pi: A \to A/A'$ is a group homomorphism.

The hint suggests that I use the same idea as in the proof of the following theorem:

Let $X$ be a set with an equivalence relation $\sim$. Let $\phi: X \to Y$ be a map of sets such that $x_1 \sim x_2$ implies $\phi(x_1) = \phi(x_2)$. Then there exists a uniquely defined map $\tilde{\phi}: X/{\sim} \to Y$ such that $\phi = \tilde{\phi} \circ \pi$.

I mention this result because the attempt that I have in mind does not proceed similarly to this proof. Here is what I had in mind.

(Attempt) Let $A' \subset A$ be normal, $\pi: A \to A/{\sim}$ the canonical quotient map. Suppose that we have defined a group structure on $A/{\sim}$ such that $\pi$ is a group homomorphism. Let $\bullet$ denote the group operation on $A$ and $\star$ denote the group operation on $A/{\sim}$. Then for any $a,b \in A$, we have $$ \pi(a \bullet b) = \pi(a) \star \pi(b), $$ so $$ \overline{a \bullet b} = \overline{a} \star \overline{b}. $$ So the group operation $\star$ must send the product of two cosets $\overline{a}$ and $\overline{b}$ to the coset of the product $a \bullet b$, where $a \bullet b$ is some element of $A$.

I assumed that there is in fact such a group structure, but I believe I also need to show that this actually gives rise to a group.

Closure. If $\overline{a}, \overline{b} \in A/{\sim}$, then $\overline{a} \star \overline{b} = \overline{a \bullet b} \in A/{\sim}$ since $a \bullet b \in A$.

Having shown that this is in fact a binary operation, I'm going to drop explicit mention of $\star$ and $\bullet$ going forward.

Associativity. Let $\overline{a}, \overline{b}, \overline{c} \in A/{\sim}$. Then we have: $$ (\overline{a} \overline{b})(\overline{c}) = \overline{ab} \overline{c} = \overline{(ab)c} = \overline{a(bc)} = \overline{a} \overline{bc} = \overline{a} (\overline{b} \overline{c}), $$ wherein we have used associativity in the group $A$.

Identity Let $e$ denote the identity in the group $A$. I claim that the coset $\overline{e}$ is the identity element we seek for $A/{\sim}$. Let $\overline{a} \in A/{\sim}$. Then $$ \overline{a} \overline{e} = \overline{ae} = \overline{a} $$ and $$ \overline{e} \overline{a} = \overline{ea} = \overline{a}, $$ so $\overline{e}$ is the desired identity element.

Inverses. Given $\overline{a} \in A/{\sim}$, $a \in A$, so $a^{-1} \in A$. Then $\overline{a^{-1}} \in A/{\sim}$. I claim that $\overline{a^{-1}} = (\overline{a})^{-1}$. We have: $$ \overline{a} \overline{a^{-1}} = \overline{aa^{-1}} = \overline{e} $$ and $$ \overline{a^{-1}} \overline{a} = \overline{a^{-1} a} = \overline{e}. $$ Therefore, $A/{\sim}$ with the proposed operation is a group.

How does this proof look?