Characteristic not implies powering-invariant in solvable Lie ring

Statement
It is possible to have a solvable Lie ring $$L$$ and a characteristic Lie subring $$M$$ of $$L$$ such that $$M$$ is not a powering-invariant Lie subring of $$L$$, i.e., there is a prime $$p$$ such that $$L$$ is $$p$$-powered (division by $$p$$ is unique in $$L$$) but $$M$$ is not $$p$$-powered.

Stronger statement for specific prime sets
Suppose $$\pi_1$$ and $$\pi_2$$ are sets of primes with $$\pi_2 \subset \pi_1$$ (as a proper subset). Then, it is possible to have a solvable Lie ring $$L$$ and a characteristic Lie subring $$M$$ of $$L$$ such that $$L$$ is powered precisely over the prime set $$\pi_1$$ and $$M$$ is powered precisely over the prime set $$\pi_2$$.

Related facts

 * Characteristic not implies powering-invariant in solvable group

Proof of the weak statement
Let $$L$$ be the two-dimensional $$\mathbb{Q}$$-Lie algebra obtained as a semidirect product of $$\mathbb{Q}$$ by $$\mathbb{Q}$$, where the latter acts on the former by multiplication in $$\mathbb{Q}$$. On account of being a $$\mathbb{Q}$$-Lie algebra, this is powered over all primes.

In this Lie algebra, the subring $$M = \mathbb{Q} \rtimes \mathbb{Z}$$ is a characteristic subring that is not powered over any prime.

Proof of the strong statement
Let $$L$$ be the two-dimensional $$\mathbb{Z}[\pi_1^{-1}]$$-Lie algebra obtained as a semidirect product of $$\mathbb{Z}[\pi_1^{-1}]$$ by $$\mathbb{Z}[\pi_1^{-1}]$$, where the latter acts on the former by multiplication in $$\mathbb{Z}[\pi_1^{-1}]$$. On account of being a $$\mathbb{Z}[\pi_1^{-1}]$$-Lie algebra, this is powered over all primes in $$\pi_1$$.

Let $$M$$ be the subring $$\mathbb{Z}[\pi_1^{-1}] \rtimes \mathbb{Z}[\pi_2^{-1}]$$. Then, $$M$$ is characteristic in $$L$$, and is powered precisely over the primes in $$\pi_2$$.