Semisimplicity of Lie algebras and non-degeneracy of associated bilinear forms of representations

Let $$\mathfrak{g}$$ be a finite-dimensional Lie algebra over a field of characteristic $$0$$. For a finite-dimensional representation $$\rho$$, we define a bilinear form $$B_\rho$$ on $$\mathfrak{g}$$ by $$B_{\rho}(x, y) = \operatorname{Tr}(\rho(x) \circ \rho(y)) \qquad (x, y \in \mathfrak{g}).$$ $$B_\rho$$ is called the bilinear form associated to $$\rho$$. Note that the Killing form is the bilinear form associated to the adjoint representation.

If $$\mathfrak{g}$$ is semisimple, $$B_\rho$$ is non-degenerate for every finite-dimensional faithful representation $$\rho$$ of $$\mathfrak{g}$$ (Bourbaki, Lie Groups and Lie Algebras, I.6.1, Proposition 1).

Question. Does the converse of the above statement hold? That is, if $$B_\rho$$ is non-degenerate for every finite-dimensional faithful representation $$\rho$$ of $$\mathfrak{g}$$, can we conclude that $$\mathfrak{g}$$ is semisimple?

(If in addition we assume that $$\mathfrak{g}$$ has a trivial center, the adjoint representation $$\mathfrak{g}$$ is faithful, and hence $$\mathfrak{g}$$ is semisimple by the assumption and Cartan’s criterion.)

$$\DeclareMathOperator\g{\mathfrak{g}}\DeclareMathOperator\h{\mathfrak{h}}$$The first preliminary remark is that $$\mathfrak{z}(\g)\cap [\g,\g]$$ is in the kernel of $$B_\rho$$ for every rep $$\rho$$. Indeed, for this we can pass to an algebraic closure, block-trigonalize the representation with irreducible diagonal blocks: on each (irreducible) diagonal block, this central intersection acts by scalars of trace zero, hence by zero.
So, combined with Ado's theorem, the assumption ensures that $$\mathfrak{z}(\g)\cap [\g,\g]=0$$. Se can write $$\g$$ as $$\mathfrak{z}(\g)\times\h$$. By Ado's theorem choose a faithful rep for $$\h$$, and take the direct sum with a faithful rep for the central factor so that it acts by strict upper triangular matrices. The direct sum of those reps is a faithful rep of the direct product, with degenerate associated bilinear form unless $$\mathfrak{z}(\g)=0$$.
Hence, $$\g$$ has trivial center, and this case is already mentioned by the OP, using the adjoint representation.