Lie bracket of a Lie ring and a subring need not be an ideal

Statement
Suppose $$L$$ is a Lie ring and $$S$$ is a subring of $$L$$. Consider the Lie bracket $$[L,S]$$, i.e., the subring of $$L$$ generated by $$[l,s]$$ for $$l \in L, s \in S$$. This need not be an ideal of $$L$$.

Proof
Let $$F$$ be a prime field for a prime $$p > 2$$, and consider the Lie ring $$L = sl(2,F)$$ of $$2 \times 2$$ matrices over $$F$$ with trace zero, with the Lie bracket given by the commutator. This is a three-dimensional vector space over the field of $$p$$ elements, and is generated by:

$$e = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}, \qquad f = \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}, \qquad h = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$.

We have:

$$[e,f] = -[f,e] = h, \qquad [h,e] = -[e,h] = 2e, [f,h] = -[h,f] = -2f$$.

Let $$S$$ be the subring of $$L$$ generated by $$e$$. In other words, $$S$$ is the cyclic subgroup generated by $$e$$. Then, $$[L,S]$$ is generated by $$[e,e], [f,e], [h,e]$$, hence it is generated as a Lie ring by $$h$$ and $$e$$. But the additive subgroup generated by $$h$$ and $$e$$ is a Lie subring, so $$[L,S] = \langle h,e \rangle$$.

On the other hand, $$\langle h,e \rangle$$ is not an ideal, because it is not closed under the Lie bracket with $$f$$.