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.  [I'll give a proof of this fact at the end]

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 $c_\beta=0$, QED.

Proof of the classification of closed G-orbits on $\mathbb{P}(\mathfrak{g})$: If x is in a closed G-orbit, then its stabiliser contains a Borel B. To be stable under the torus T implies that x lies in a single weight-space. To be stable under the unipotent radical implies that that weight space must be the highest weight. (an alternative approach to this result is the argument in the proof of Theorem 4.3.3 in Collingwood and McGovern's "Nilpotent Orbits in Semisimple Lie Algebras").