Here's what I need to show:
Consider the action of $G$ on itself by conjugation. In other words, if $g,a \in G$, then the action of $g$ on $a$ is defined by $g*a = gag^{-1}$. Then show that if $G$ is not an abelian group, there is an orbit of order at least $2$.
Here's my attempt:
Suppose that $\mathcal{O} _{s} =\{s\}$ for all $s\in G$. Thus, we must have $gsg^{-1}=s$ for all $g,s\in G$. Let $a,b \in G$ be arbitrary. Then $a=bab^{-1}$ and $b=aba^{-1}$. Now, $$\begin{align} ab &= (bab^{-1})(aba^{-1}) \\ &= ba(b^{-1}ab)a^{-1} \\ &= ba(bab^{-1})a^{-1} \\ &=baaa^{-1} \\ &=ba. \end{align}$$ Since $a,b \in G$ were arbitrary, this means $G$ is abelian which contradicts our hypothesis.
Is this proof okay?
It seems like you might want to indicate why $\mathcal O_s=\{s\}\,,\forall s\in G$, other than just that it has order $1$. I know it's trivial, but perhaps mention that always $s\in\mathcal O_s$, because of the fact that $e*s=s\,,\forall s$. (I understand it's a general fact about group actions.) However, it's up to you.
On the other hand, when $gsg^{-1}=s\,,\forall g,s\in G$, we immediately have that $gs=sg$, and hence the group is abelian.
Finally, in the middle of your proof, I noticed that $b^{-1}ab$ gets replaced by $bab^{-1}$. This might not be true. To correct it, note that $b^{-1}ab=b^{-1}a{(b^{-1})}^{-1}=a$, by assumption.