I will sketch the proof that over the complex numbers, the answer is no.
The set $$\{x\in \mathbb{P}(\mathfrak{g}) \mid \exists y\neq 0, [[\mathfrak{g},x],y]=0\}$$ is closed and $G$-invariant. Therefore it suffices to assume that x lies in a closed $G$-orbit in $\mathbb{P}(\mathfrak{g})$.
So we can assume that $x\in \mathfrak{g}_\alpha$ for some root $\alpha$, where we have also fixed a Cartan subalgebra $\mathfrak{h}$ to talk about root spaces.
Now write $y=h+\sum_\beta c_\beta X_\beta$, where $h\in \mathfrak{h}$ and $X_\beta\in \mathfrak{g}_\beta$. Then $[y,X_\gamma]=0$ for all $\gamma$ with $\gamma-\alpha$ a root or zero.
The set of possible $\gamma$ does not lie in a hyperplane, which forces $h=0$. For every root $\beta$ there exists such a $\gamma$ with $\gamma+\beta$ a root or zero, which forces $c_\beta=0$.
