 Solution

Let $A$ be the two-dimensional affine Lie algebra with basis $u,v$ and bracket $[u,v]=v$, and putBoth factors have zero centre, so $Z(L)=0$. The line $\mathbb{C}v$ in the second factor is an ideal, making $L$ PSS. But its quotient onto the nonabelian simple algebra ${\mathfrak{sl}}_2$ is not soluble. A quotient of a soluble algebra would be soluble, so $L$ is not soluble.