Finite groups with 3 conjugacy classes

Let $G$ act on $X:=G$ by conjugation, that is via the action $(g,x)\in G\times X\longmapsto g\cdot x=gxg^{-1}\in G$.

Thus the orbit of $x$ under this action is $O_x:=\{gxg^{-1}, g\in G\}$, i.e., the conjugacy class of $x$, and its stabilizer is $Stab(x):=\{g\in G/ gxg^{-1}=x\}$, i.e., the set of elements that commute with $x$.

We know that the set of orbits forms a partition of $X$, that is, there exist $x_1,\dots,x_r\in X$ such that $X=\bigsqcup_{i=1}^r O_{x_i}$. But there are only three conjugacy classes here, so $r=3$. Also, one of these classes contains the identity element $e$, whose orbit is reduced to itself since it commutes with everyone. We thus have the equality $X=\{e\}\sqcup O_x\sqcup O_y$ for some $x,y\in X$.

Let us now consider cardinalities: we find $\#G=1+\#O_x+\#O_y$. But the cardinalities $\#O_x$ and $\#O_y$ can also be written as $\frac{\#G}{\#Stab(x)}$ (samely for $y$), so we finally obtain
$$\#G=1+\#G\left(\frac{1}{\#Stab(x)}+\frac{1}{\#Stab(y)}\right).$$

Which can be rewritten as
$$\#Stab(x)\cdot\#Stab(y)\cdot(\#G-1)=\#G\cdot(\#Stab(x)+\#Stab(y))$$But since $\#G$ and $\#G-1$ have gcd $1$, then $\#G-1$ divides $\#Stab(x)+\#Stab(y)$. However $Stab(x)$ and $Stab(y)$ are subgroups of $G$. Say that one of them equal $G$, then $\#G-1$ divides the cardinality of the other one minus one, and so it also has to equal $G$. But $Stab(x)=G$ iff $x$ commutes with any element of $G$ and samely for $y$. Hence $G=\Z/3\Z$.

Now assume that both $Stab(x)$ and $Stab(x)$ do not equal $G$. Then, by Lagrange formula, their cardinal has to be less than or equal to $\frac{\#G}{2}$, so their sum has to be less than $\#G$ and is a multiple of $\#G-1$. Hence it has to equal $\#G-1$.
Finally one gets the two equations:
$$\#G-1=\#Stab(x)+\#Stab(y)\,\,\text{and}\,\,\#G=\#Stab(x)\cdot\#Stab(y).$$

Since we know that $e\in Stab(x)\cap Stab(y)$, let us write $a:=\#Stab(x)-1$ and $b:=\#Stab(y)-1$, which gives $a+b=\#G-3$ and $2=ab$. Hence, without loss of generality, one deduces that $a=2$ and $b=1$ hence $\#Stab(x)=3,\#Stab(y)=2$ and $\#G=6$.

Finally, there exists only one non-abelian group of order $6$ (it has to be non-abelian to have only $3$ conjugacy classes), which is $S_3$. This concludes the proof.