Finite groups with 2 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 two conjugacy classes here, so $r=2$. 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 $X=\{e\}\sqcup O_x$ for some $x\in X$.

Let us now consider cardinalities: we find $\#G=1+\#O_x$. But the cardinality $\#O_x$ can also be written as $\frac{\#G}{\#Stab(x)}$, so we finally obtain $\#Stab(x)=\frac{\#G}{\#G-1}$, which must be an integer. This is only possible if $\#G=2$.

Conversely, if $G$ contains only two elements, it indeed contains only two conjugacy classes since it is abelian, which concludes the proof.