This is not a complete answer, but there is a nice description of the information in $P_T$ which may prove useful to someone else.

First of all, I will define $P_T(z)$ slightly differently: If $T'$ is the leafing of $T$, then I define $P_T'(z)= zP(z)+1$.  Since this is just your polynomial evaluated at $z-1$, it determines $T$ just as well as yours.

Then **the coefficient of $z^n$ in $P_T(z)$ is the number of $n$-node subtrees of $T$** (i.e. a subgraph of $T$ containing the root and $n$ other vertices).  This is because, for $n>0$, choosing an $n$-node subtree of the leafing of $T$ is the same as choosing an $(n-1)$-node subtree of $T$, and for any $n$, choosing an $n$-node subtree of the grafting of $T$ and $T'$ is the same is choosing a $k$-node subtree of $T$ and an $(n-k)$-node subtree of $T'$ for some $k$ between $0$ and $n$.  

Some consequences include:

 - If $T$ has $n$ nodes (vertices other than the root), then the highest-order term of $P_T(z)$ is $z^n$.
 - The coefficient of $z$ in $P_T(z)$ is the degree of the root of $T$.
 - If $T$ has $a$ nodes at distance $1$ from the root, and $b$ nodes at distance $2$, then the coefficient of $z^2$ in $P_T(z)$ is ${a \choose 2}+ b$.
 - If $T$ has a total of $n$ nodes, then the coefficient of $z^{n-1}$ is the number of leaves of $T$ (nodes with degree 1).

Hence if $P_T(z)=P_{T'}(z)$, then $T$ and $T'$ have the same numbers of vertices and leaves, their roots have the same degrees, and they have the same total number of vertices at distance $2$ from the root.  It seems that more should be true, but I haven't proven any more.