Merkle tree alternating hash and polynomial

I want to get feedback on the security of a modified merkle tree data structure. Using the image above as a reference assume I have a random oracle function $H$. Assume $H$ outputs a value in $\mathbb{F}_p$, and all math is happening in this field too.

The standard approach to calculate $D$ is

$B = H(A, C)$

$F = H(E, G)$

$D = H(B, F)$

The approach I'm considering alternates between hash function and polynomial. To calculate $D$ I would use $H$ in the first level:

$B = H(A, C)$

$F = H(E, G)$

Then use multiplication in the second level:

$D = B * F$

Then to calculate the parent of $D$ I would do $H(D, D_{sibling})$, then above that I would use multiplication again.

I'm wondering, is this guarded by Schwartz–Zippel? $B$ and $F$ should both be randomly distributed in $\mathbb{F}_p$, is $B*F$ a polynomial of degree 2? Does the case of $B = F$ matter? e.g. $A = E \land C = G$

When making a merkle proof I'd include the pre-image of any nodes that are combined with multiplication. So in the below tree I would prove membership of $J$ in $A$ by supplying: $J$, $K$, $I$, $Z$, $C$.

          A
       /      \
      B        C
     /  \     /  \
    D    E   F    G
   / \  / \ / \  / \
  Z  I J  K L  M N  O

I would then compute

$B = H(Z, I) * H(J, K)$

$A = H(B, C)$

I want to get feedback on the security of a modified merkle tree data structure.

Doesn't look good, at least as well as using it to prove membership within the Merkle tree.

Here's the issue: for multiplication, you can compute preimages. You don't say what ring you compute the multiplication over, but for anything reasonable, given a target $C$ and several possible values $A_1, A_2, ...$, you can usually find a $B$ for which $A_i \times B = C$ for some $i$ (and if it's a finite field, unless $A_1 = 0$, only one value is needed).

Here's why that's an issue: suppose you see a Merkle tree proof that $A$ is a part of the Merkle tree; that proof consists of the values of $B$ and $F$. We want to prove that $A'$ is a part of the Merkle tree; we generate several values $B_1, B_2,...$, compute $H(A, B_i)$, and for that set of values, find an $F'$ such that $H(A, B_i) \times F' = D$; and we're done.