This is indeed the case. To see this, filter $\mathfrak{g}$ by $Ker(ad_e^i)$. Let $Gr(\mathfrak{g})$ be the associated graded. We have a symbol map $sym : \mathfrak{g} \to Gr(\mathfrak{g})$ given by sending an element to the "leading term" w.r.t. this filtration. It would suffice to show that it is generated by the symbols of elements of $\mathfrak{g}_e + [x,\mathfrak{g}]$, by a standard descent argument (look at the least element not in your subspace in terms of this filtration) But, after passage to the associated graded, $ad_x$ acts on symbols like $ad_f$ modulo previous graded pieces. More precisely, we have $sym(ad_x(u)) = sym(ad_f(u))$ for every $u$. This shows that the image of the symbol don't change if you replace $x$ by $f$, so we can reduce to this case. But for $x = f$ the statement is a standard result on representations of $\mathfrak{sl}_2$ with no trivial components (which follows from the triple being principal in this case).
Edit: As mensioned in the comment below, the condition on non-triviality of the representation is irrelevant here. It is just true for every $\mathfrak{sl}_2$ representation.