There are various ways to generalize the equation. I give two different ones here:
(1) In terms of bimodules: Take a finite index irreducible subfactor $N \subset M$. Suppose $P_1$ and $P_2$ are two intermediate subfactors and $P=P_1 \cap P_2$. Then we have the following equation for $N-N$ bimodules:
$$\dim(P_1\mathbin{\mathop{\otimes}\limits_{P}} P_2)\dim(P)=\dim(P_1)\dim(P_2).$$
Proof: \begin{align*}
\dim(P_1)\dim(P_2)&=\dim(P_1\mathbin{\mathop{\otimes}\limits_{N}}P_2)\\
&=\dim(P_1\mathbin{\mathop{\otimes}\limits_{P}}P\mathbin{\mathop{\otimes}\limits_{N}}P\mathbin{\mathop{\otimes}\limits_{P}}P_2)\\
&=\dim(P_1\mathbin{\mathop{\otimes}\limits_{P}} P_2)\dim(P\mathbin{\mathop{\otimes}\limits_{P}}P)\\
&=\dim(P_1\mathbin{\mathop{\otimes}\limits_{P}} P_2)\dim(P).
\end{align*}
When $M=N\rtimes G$, $P_i=N\rtimes H_i$, for $i=1,2$, one obtains the group case.
(2) In terms of two boxes in planar algebras:
We have an inequality in your formulation.
Proof: Let $b,b_1,b_2$ be biprojections corresponding to $P,P_1,P_2$ in case (1).
Then
\begin{align*}
tr(S(b_1*b_2))&=\dim(P_1P_2)\leq\dim(P_1\mathbin{\mathop{\otimes}\limits_{N}}P_2);\\
tr(b)&=\dim(P);\\
tr(b_1)&=\dim(P_1);\\
tr(b_2)&=\dim(P_2).
\end{align*}
By the equation in case (1), we have that
$$tr(S(b_1*b_2))tr(b)\leq tr(b_1)tr(b_2).$$
The equality holds iff $b_1b_2=b$. This condition always holds for commutative algebras, but not for non-commutative algebras.
Instead we can prove the following equality using the exchange relation of biprojections.
$$tr(S(b_1*b_2))tr(b_1b_2)=tr(b_1)tr(b_2).$$
(I used this formula to show that the lower bound of the angle between minimal biprojections is $60^\circ$.)