Since $B(z)$ is an exponential generating function $B(z) = \sum_n b_n \frac{z^n}{n!}$, and not $\sum_n b_n z^n$.
With correct definition the series expansion for $B(z)$ should read:
$$
B(z) \sim z+\frac{z^2}{2}+\frac{z^3}{2}+\frac{5 z^4}{8}+\frac{7 z^5}{8}+\frac{21
z^6}{16}+\frac{33 z^7}{16}+O\left(z^8\right)
$$
and this satisfies the recurrence equation:
In[200]:= With[{b = Sum[(2 n - 3)!!/n! z^n, {n, 1, 8}] + O[z]^8},
z + 1/2 b^2 - b]
Out[200]= SeriesData[z, 0, {}, 8, 8, 1]
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.
Best Answer
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...