$\DeclareMathOperator\g{\mathfrak{g}}\DeclareMathOperator\h{\mathfrak{h}}$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.
(Note that the result trivially implies Ado's theorem, so we can't expect to bypass it.)
It's enough that the field characteristic is $\ne 2$.
Let $\beta$ be an invariant symmetric bilinear form on $\mathfrak{sl}(2)$.
\begin{align}
0 &= \beta(h, 0) = \beta(h, [x,x]) = \beta([h, x], x) = \beta(2x, x)\\
0 &= \beta(h, 0) = \beta(h, [y,y]) = \beta([h, y], y) = \beta(-2y, y)\\
0 &= \beta(y, 0) = \beta(y, [x,x]) = \beta([y, x], x) = \beta(-h, x)\\
0 &= \beta(x, 0) = \beta(x, [y,y]) = \beta([x, y], y) = \beta(h, y)\\
\beta(2x, y) &= \beta([h, x], y) = \beta(h, [x,y]) = \beta(h, h)
\end{align}
So $\beta(x,x) = \beta(y,y) = \beta(h, x) = \beta(h, y) = 0$ and $\beta(h,h) = 2\beta(x,y)$. Ordering the basis as $(x, h, y)$ and writing in a matrix, the matrix of $\beta$ is a scalar multiple of
\begin{pmatrix}
0 & 0 & 1\\
0 & 2 & 0\\
1 & 0 & 0
\end{pmatrix}
and in fact, this is a valid invariant symmetric binlear form. It's the one associated with $\text{tr}(\phi(x)\phi(y))$ where $\phi$ is the representation of $\mathfrak{sl}(2)$ on a two dimensional space.
Best Answer
Any trace form is invariant because of ${\rm tr}([A,B]C)={\rm tr}(A[B,C])$ for $A,B,C \in {\rm End}(V)$.
Indeed, using Jyrki's comment with ${\rm tr}(BAC)={\rm tr}(ACB) $ we have $$ {\rm tr}([A,B]C)={\rm tr}((AB-BA)C)={\rm tr}(ABC-BAC)={\rm tr}(ABC-ACB)={\rm tr}(A[B,C]). $$