Quotient Group G/G = {identity}?

Well, forgive me for getting all technical, but this isn't right: $G/G = \{ g \in G \colon gG\}$. It should be the other way around: $G/G = \{ gG |\, g \in G \}$.

Next, we get rid of $g$ by realizing that $gG = G$ for any $g \in G$. Therefore, $G/G = \{G\}$, i.e. $G/G$ is a set with exactly 1 element, and this element is $G$ itself.

Now, the next thing is to realize how the operation in $G/G = \{G\}$ works. It works like this: $G \cdot G = G$. Exactly like that of the trivial group $\{1\}$: $1 \cdot 1 = 1$. So, the map $G/G \to \{1\}$ that sends $G$ to $1$ is a group isomorphism.

I know, I know, this may be the worst way to explain these things. There are way too many trivial formulas there. But it can be good for convincing oneself of something.

$\operatorname{id}_G$ is not an element of $G/G$. The elements of $G/G$ are sets of the form $gG$ with $g\in G$. But $gG=G$ for all $g\in G$, so there is only one co-set.

Now you just have to convince yourself that all groups with only one element are isomorphic. That's pretty easy to do.

Consider the zero homomorphism $\phi:G\rightarrow G$. Then the $\ker\phi=G$, $\mathrm{Im}\phi =\left\{id_G\right\}$ and from the first theorem of Isomorphism $\frac{G}{\ker\phi}\cong \mathrm{Im}\phi$ which of course implies that $G/G\cong \left\{id_G\right\}$