Use the analytic method. Your class is a root connected to a non-empty set of trees, or a leaf. Use $\mathcal{Z}$ (and $z$) for inner nodes, $\mathcal{Y}$ (and $y$) for leaves; use $\mathcal{E}$ for the class with one empty object:
$$
\mathcal{T}
= \mathcal{Z} \star (\mathfrak{S}(\mathcal{T}) \smallsetminus \mathcal{E})
+ \mathcal{Z} \mathcal{Y}
$$
This translates to:
$$
T(z, y) = z (e^{T(z, y)} - 1) + z y
$$
Just need to get $T(z, y)$ (or the coefficients) out of this...
The connection between the generating functions for unlabeled trees and for unlabeled rooted trees is derived in The Number of Trees, Richard Otter, The Annals of Mathematics, $2$nd Ser., Vol. $49$, No. $3$ (Jul., $1948$), pp. $583$-$599$ (in the section titled “Trees” starting on p. $587$).
The proof hinges on the fact that if you form equivalence classes of the vertices and edges of a tree according to whether they are transformed into each other by isomorphisms of the tree, you can choose representatives that form a subtree, with the exception of what Otter calls a “symmetry edge”, an edge which is flipped by an isomorphism of the tree and forms an equivalence class of its own. Since the number of vertices in the subtree is one more than the number of edges, it follows that the number $t_n$ of unlabeled trees is the number $a_n$ of unlabeled rooted trees minus the number $b_n$ of unlabeled “edge-rooted” trees (i.e. unlabeled trees with a distinguished edge, which must not be a symmetry edge). The number of edge-rooted trees can be counted by counting the ways of splitting trees at an edge and rooting the two resulting trees at the vertices of the edge:
$$
b_n=\frac12\left(\sum_{k=0}^na_ka_{n-k}-a_{n/2}\right)\;,
$$
where the factor $\frac12$ accounts for double-counting, $a_{n/2}=0$ if $n$ is odd, and the second term subtracts the contribution from symmetry edges. Taking into account the special case $n=0$, where we have the empty tree but no rooted empty tree, we have
$$
t_n=\delta_{n0}+a_n-b_n\;,
$$
and translating this to generating functions yields
$$
T(x)=1+A(x)-\frac12\left(A(x)^2-A(x^2)\right)\;.
$$
Best Answer
We will compute the number of unlabeled ordered rooted trees on $n$ nodes and having $k$ leaves.
The combinatorial class equation for these trees with leaves marked is
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \mathcal{T} = \mathcal{Z}\times\mathcal{U} + \mathcal{Z} \times \textsc{SEQ}_{\ge 1}(\mathcal{T}) \quad\text{or}\quad \mathcal{T} = \mathcal{Z}\times\mathcal{U} + \mathcal{Z} \times \sum_{p\ge 1} \mathcal{T}^p.$$
This yields the functional equation for the generating function $T(z)$ $$T(z) = zu + z\frac{T(z)}{1-T(z)}$$ or $$z = \frac{T(z)}{u+T(z)/(1-T(z))} = \frac{T(z)(1-T(z))}{T(z)+u(1-T(z))}.$$
Note that leaves in addition to being marked as such also carry the node marker so that the total number of nodes includes the leaves. If this is not desired subtract the number of leaves from the number of nodes to get the count of genuine internal nodes.
Starting the computation we seek
$$n T_n(u) = [z^{n-1}] T'(z) = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n}} T'(z) \; dz.$$
and will compute this by a variant of Lagrange inversion. We put $T(z) = w$ so that $T'(z) \; dz = dw$ and we find (here we have used the fact that $w=uz+\cdots$)
$$\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(w+u(1-w))^n}{w^n(1-w)^n} \; dw.$$
Extract the coefficient on $[u^k]$ to get
$${n\choose k} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1-w)^k w^{n-k}}{w^n(1-w)^n} \; dw \\ = {n\choose k} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^k} \frac{1}{(1-w)^{n-k}} \; dw.$$
Collecting everything we thus have
$$[u^k] [z^n] T(z) = \frac{1}{n} {n\choose k} {k-1+n-k-1\choose n-k-1}$$
or indeed
$$\bbox[5px,border:2px solid #00A000]{ [u^k] [z^n] T(z) = [u^k] T_n(u) = \frac{1}{n} {n\choose k} {n-2\choose n-k-1}}$$
as claimed. This formula holds for $n\ge 2$ where $1\le k\le n-1.$ Note that the case $k=0$ will always produce zero as it ought to (there is no ordered tree with no leaf) owing to the binomial coefficient ${n-2\choose n-1}.$ Note however that when $n=1$ and $k=0$ we get ${-1\choose 0}$ which evaluates to one, yet the ordered tree on one node is also a leaf.
There is an earlier version of this computation at the following MSE link, which is not as streamlined yet includes a verification of the closed form using the Maple combstruct package.
Re-writing the binomial coefficients we find
$$\frac{1}{n} {n\choose k} {n-2\choose n-k-1} = \frac{1}{n} {n\choose k} {n-2\choose k-1} = \frac{1}{k} {n-1\choose k-1} {n-2\choose k-1}.$$
This choice of representation makes it clear that what we have here are Narayana numbers from the Catalan triangle, shifted by one. This is OEIS A001263. We can also prove that these values add to the Catalan numbers, shifted as well.
We get
$$\sum_{k=1}^{n-1} \frac{1}{n} {n\choose k} {n-2\choose n-k-1} = \frac{1}{n} \sum_{k=0}^{n-1} {n\choose k} {n-2\choose n-k-1} \\ = \frac{1}{n} \sum_{k=0}^{n-1} {n\choose k} [z^{n-k-1}] (1+z)^{n-2} = \frac{1}{n} [z^{n-1}] (1+z)^{n-2} \sum_{k=0}^{n-1} {n\choose k} z^k.$$
We may extend $k$ beyond $n-1$ owing to the coefficient extractor in front:
$$\frac{1}{n} [z^{n-1}] (1+z)^{n-2} \sum_{k\ge 0} {n\choose k} z^k = \frac{1}{n} [z^{n-1}] (1+z)^{n-2} (1+z)^n \\ = \frac{1}{n} [z^{n-1}] (1+z)^{2n-2} = \frac{1}{n} {2n-2\choose n-1}.$$
These are indeed the familiar Catalan numbers, thus shown to count ordered trees.