Now, our techniques to solve this combinatorics problem:
We studied an operad (the operad of posets), and then we studied power series indexed by posets.

Since the power series are indexed by posets, endomorphisms of posets can be extended to endomorphisms of those power series. That is, if we have a series $F(P)$ for every poset $P$, and we have $Q$ a function taking $P_1,...,P_n$ posets to a poset $Q(P_1,...P_n)$, then we define $Q(F(P_1),...,F(P_n)):=F(Q(P_1,...P_n))$.

If we work with sets, then we dont have to worry too much about the properties of this induced functions. They are just set theoretical functions.

Now, the correct language here is the language of #operads. Since we work with the operad of posets, this are topological operads. It is just that the topological spaces are finite but with a topology distinct from the discrete topology!.

Returning to the problem of enumerating shuffles between a tree and a linear tree. If we consider the series $F(P)$ whose $n$ coefficient enumerates shuffles between a poset $P$ and the linear order with $n$ cards, here $P$ is the poset of vertices of the input tree,
then we have an action of the operad of posets of the form $Q(F(P_1),...,F(P_n)):=F(Q(P_1,...P_n))$.
Now, the difficult part is to compute the actions of the posets. But in the paper https://doi.org/10.1007/s10801-025-01386-7 we were able to compute the action of the posets $\{x,y\}$ and $\{x<y\}$, this is enough to compute the series of any series parallel poset, which includes trees.

Sometimes, the action of the operad preserves the combinatorial / enumerative property of the series. That is, the operad sends series solving a problem to series solving the same problem. #EnumerativeCombinatorics 2/3