Showing all p-local subgroups are char p given that all are contained in char p locals

Question: How does one prove that if every $p$-subgroup $U$ is contained as a subnormal subgroup of a characteristic-$p$, radical, local subgroup containing the normalizer of $U$, then the normalizer of every $p$-subgroup $U$ has characteristic $p$?

Here a group $X$ is said to have characteristic $p$ if $F^*(X) = O_p(X)$, that is, every minimal normal subgroup is a $p$-group and there are no subnormal subgroups $N$ with $N/Z(N)$ simple.

In other words, suppose $G$ has the (weaker) property: If $U$ is a $p$-subgroup of $G$, then there is a $p$-subgroup $R \leq G$ such that $N_G(U) \leq N_G(R)$ and $U \leq R = O_p( N_G(R) ) = F^*( N_G(R) )$ where $F^*(X)$ is the generalized Fitting subgroup of $X$.

I would like to conclude that $G$ has the (seemingly much stronger) property that if $U$ is a $p$-subgroup of $G$, then $O_p(N_G(U)) = F^*( N_G(U) )$.

Motivation: I'm writing up an answer to this question and showed such groups have the weaker property. I'm reasonably confident this is a good direction, since page 94 of number 3 in GLS's CFSG claims that this follows from Thompson's A×B lemma on page 73 of number 2 in GLS's CFSG:

Lemma: (Thompson's A×B Lemma) Assume $A\times B$ acts on $P$, where $A$ is a $p'$-group and $B$ and $P$ are $p$-groups. If $A$ does not centralize $P$, then $A$ does not centralize $C_P(B)$. More generally, suppose that $AB$ acts on $P$, where $B$ and $P$ are $p$-groups and $[A,B] = 1$. If $A$ centralizes $C_P(B)$, then $O^p(A)$ centralizes $P$.

Minimal Progress: Set $L=N_G(U)$ to be the local subgroup of $G$, and $K=O_p(L)$ to be its $p$-core. We need to show $F^*(L) = K$. By the property, we can find $R$ such that $K \leq R = O_p(N_G(R)) = F^*(N_G(R))$ and $N_G(K) \leq N_G(R)$. Indeed, one can check that $N_G(U) = L \leq N_G(K) \leq N_G(R)$. However, we need control on how $N_G(U)$ sits inside $N_G(R)$ to compare $F^*(N_G(U))/O_p(N_G(U))$ to $F^*(N_G(R))/O_p(N_G(R))$.

I'm not sure how to use Thompson's A×B lemma. My best guess is the definition of the generalized Fitting as the intersection of the extended centralizers of the chief factors, but certainly not all of the chief factors are $p$-groups.

Disturbing example: So a super-easy idea (that I wrote up to near completion before I realized it was flawed) was an induction. In the specific $G$, $N_G(R)/R$ is actually isomorphic to a direct product of smaller $G$ with the property, so if direct products and quotients by $p$-core work nicely, then everything is good. Unfortunately $(G=A_5,p=2)$ has the property (both weak and strong) while (somewhat obviously) $(G=A_5 \times A_5,p=2)$ does not (have either). If $G$ is $\operatorname{PSL}(4,4)$ then $N_G(R)/R$ can be $A_5 \times A_5$, hence the quotient induction cannot work.

Lemma: If $U \leq O_p(X)$, then $O^p(F^*(N_X(U)))=O^p(F^*(X))$.
This basically finishes it. Let $U$ be a $p$-subgroup of some group $G$ with the weak property. Then find $U \leq R$ with $N_G(U) \leq N_G(R)$ and $R=F^*(N_G(R))$. Set $X=N_G(R)$. Note that $N_X(U) = X \cap N_G(U) = N_G(U)$. Then $$O^p(F^*(N_G(U))) = O^p(F^*(N_X(U))) = O^p(F^*(X)) = O^p(R) = 1,$$ so $F^*(N_G(U))$ is a $p$-group as required.