I already proved that $a \sim b$ is an equivalence relationship.
I think that identity element $e$ of $G$ is defined as: $b b^{-1}$ because $(ab^{-1})(bb^{-1}) = ab^{-1}$
Now I'm just realizing that I also have to prove that $e\ a = a$.
so $(aa^{-1})(ab^{-1}) = ab^{-1}$ as well.
so either $a a^{-1}$ or $b b^{-1}$ are both identity elements?
I'm also getting caught up on what an equivalence class for an identity element actually is.
On
A bit to unpack here.
First, the identity element $e$ of $G$ is not defined by $bb^{-1}$. It is defined as the (unique) element $e$ such that $ae=ea = a$ for all $a \in G$. This is the definition, so don't have to prove that $ea = a$. As it happens, $aa^{-1}$ and $bb^{-1}$ are both the identity element (by definition of inverse elements).
So, what is the equivalence class of an element $a \in G$? By definition, it is the set of all $b \in G$ such that $a \sim b$. So, the equivalence class of $e \in G$ (the identity element) is the set of all $b$ such that $e \sim b$. Unwrapping the definition of $\sim$ tells us that the equivalence class we're looking for is the set of all $b$ such that $eb^{-1} = b^{-1} \in H$. But, $b^{-1} \in H$ if and only if $b \in H$, so the equivalence class of $e$ is just $H$.
On
One way to view $a\sim_H b$ is to see it as equivalent to saying $a$ and $b$ differ by an element in $H$ with respect to the operation of the group. This is most clear when the operation is addition, since then $a\sim_H b$ iff $a-b\in H$.
As such, then, the equivalence class of $e$ is all $g\in G$ such that $g$ and $e$ differ by an element of $H$ (and I can write $g$ first as $\sim_H$ is symmetric as an equivalence relation); namely, $g\sim_H e$ iff $g=ge^{-1}\in H$.
Hence the equivalence class is
$$\begin{align} [e]_{\sim_H}&:=\{ g\in G\mid e\sim_H g\}\\ &=H. \end{align}$$
Since $a \sim e \Leftrightarrow ae^{-1}=a \in H$, the equivalence class of $e$ is $H$. More generally, the equivalence class of $b\in G$ is the right coset $Hb$.