If a group $G$ contains an element a having exactly two conjugates, then $G$ has a proper normal subgroup $N \ne \{e\}$ If a group $G$ contains an element a having exactly two conjugates, then $G$ has a proper normal subgroup $N \ne \{e\}$
So my take on this is as follows: If we take $C_G(S)$ of S. This is a subgroup of G. If $C_G(S)=G$, then S has no conjugate but itself, so therefore $C_G(S)$ is a proper subgroup. If we suppose $C_G(S)=\{e\}$,then in order for there to be exactly two conjugates of S, then
For every $a \ne b \in G \ \{e\}, bxb^{-1} =axa^{-1}$ but $bxb^{-1}=axa^{-1} \to (a^{-1}b)xb^{-1}=xa^{-1} \to (a^{-1}b)x(a^{-1}b)^{-1}=x \to a^{-1}b \in C_G(S)$
Which means that $C_G(S)$ is actually nontrivial or that $a^{-1}b=e$ if and only if $a=b$, which would be a contradiction. Thus $C_G(S)$ is a nontrivial proper subgroup. Since there are exactly 2 conjugacy classes of S and they are in one to one correspondence with cosets of S, its centralizers' index $[G:C_G(S)]=s$. Subgroups of index $2$ are normal, so $C_G(S)$ is a proper nontrivial normal subgroup.
This approach seemed very different from other examples I have seen so I guess I am wondering if this approach makes sense.

Simplifying $a,b$ argument a little:
Let $g$ be the element with exactly two conjugates. Suppose $C_G(g) = \{e\}$. Since $g \in C_G(g)$, this means $g = e$. 
By Lagrange's Theorem, $[G : C_G(g)] = 2$ imples $\lvert G \rvert = 2$, so $G = \{e, h\}$. The two conjugates of $g$ are itself and $hgh^{-1}$, but both are $e$, contradicting $g$ having two conjugates. 
Conclude $C_G(g) \ne \{e\}$. 
