Here is some enrichment material to complete this calculation. I
assume you are working with labeled trees since you mention an
exponential generating function. First note that these planted trees
correspond to ordinary rooted trees with an extra node attached at the
root.
The combinatorial class equation here is
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\mathcal{T} =
\mathcal{Z} \times \textsc{SET}(\mathcal{T}).$$
This gives the functional equation
$$T(z) = z \exp T(z)$$
from which Lagrange inversion produces
$$n! [z^n] T(z) = n^{n-1}.$$
Now if we wish to mark nodes with an even number of children
we get
$$\mathcal{T} =
\mathcal{Z} \times \mathcal{U} \times \textsc{SET}_\mathrm{even}(\mathcal{T})
+ \mathcal{Z} \times \textsc{SET}_\mathrm{odd}(\mathcal{T}).$$
Introduce $Q(z, u)$ for this specification so that $Q(z,1) = T(z).$ We
obtain the functional equation
$$Q(z) = uz \sum_{k\ge 0} \frac{Q(z)^{2k}}{(2k)!}
+ z \sum_{k\ge 0} \frac{Q(z)^{2k+1}}{(2k+1)!}.$$
Now to count the number of nodes with an even number of children
introduce
$$G(z) = \left. \frac{\partial}{\partial u} Q(z, u) \right|_{u=1}.$$
Differentiate the functional equation with respect to $u$ and put
$u=1$ to obtain
$$G(z) = z \sum_{k\ge 0} \frac{T(z)^{2k}}{(2k)!}
+ z \sum_{k\ge 1} \frac{T(z)^{2k-1}}{(2k-1)!} G(z)
+ z \sum_{k\ge 0} \frac{T(z)^{2k}}{(2k)!} G(z).$$
This yields
$$G(z) = z \sum_{k\ge 0} \frac{T(z)^{2k}}{(2k)!}
+ z \times G(z)\exp T(z)$$
which is
$$G(z) = \frac{1}{2} z (\exp T(z) + \exp(-T(z)) + G(z) T(z)$$
or $$G(z) = \frac{1}{2} T(z) + \frac{1}{2} \frac{z^2}{T(z)}
+ G(z) T(z).$$
This finally produces
$$G(z) = \frac{T(z)^2+z^2}{2T(z)(1-T(z))}.$$
Now to extract coefficiets from this we use the Cauchy Coefficient Formula.
Start from
$$n! [z^n] G(z) =
\frac{n!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} \frac{T(z)^2+z^2}{2T(z)(1-T(z))} \; dz.$$
There are two components here.
First component.
Put $T(z)=w$ so that $z=w/\exp(w) = w\exp(-w)$
and $dz = (\exp(-w) - w\exp(-w)) \; dw$
to get
$$\frac{n!}{2\pi i}
\int_{|w|=\epsilon}
\frac{\exp(w(n+1))}{w^{n+1}}
\times
\frac{w^2}{2w(1-w)}
\times (\exp(-w) - w\exp(-w)) \; dw
\\ = \frac{1}{2} \frac{n!}{2\pi i}
\int_{|w|=\epsilon}
\frac{\exp(wn)}{w^n} \; dw
= \frac{1}{2} n! \frac{n^{n-1}}{(n-1)!}
= \frac{1}{2} n^n.$$
Second component.
Using the same substitution we obtain
$$\frac{n!}{2\pi i}
\int_{|w|=\epsilon}
\frac{\exp(w(n+1))}{w^{n+1}}
\times
\frac{w^2\times \exp(-2w)}{2w(1-w)}
\times (\exp(-w) - w\exp(-w)) \; dw
\\ = \frac{1}{2} \frac{n!}{2\pi i}
\int_{|w|=\epsilon}
\frac{\exp(w(n-2))}{w^n} \; dw
= \frac{1}{2} n! \frac{(n-2)^{n-1}}{(n-1)!}
= \frac{1}{2} n\times (n-2)^{n-1}.$$
This gives the following sequence:
$$1, 2, 15, 144, 1765, 26400, 466459, 9508352,
\\ 219651849, 5671088640, \ldots $$
Conclusion. The average number of even outdegree nodes
among the labeled trees on $n$ nodes is
$$\frac{1}{n^{n-1}} \times
\frac{1}{2} \left(n^n + n \times (n-2)^{n-1}\right)$$
which yields
$$\frac{1}{2} n
+ \frac{1}{2} \frac{(n-2)^{n-1}}{n^{n-2}}.$$
To get the asymptotics re-write this like so:
$$\frac{1}{2} n
+ \frac{1}{2} \times n \times \frac{(n-2)^{n-1}}{n^{n-1}}
= \frac{1}{2} n
+ \frac{1}{2} \times n \times \left(1-\frac{2}{n}\right)^{n-1}.$$
This gives
$$\frac{1}{2} \times n \times \left(1+\frac{1}{e^2}\right).$$
There is a bias towards even-degree nodes.
This was also seen at this
MSE link
where unlabeled trees were studied.
Remark. These data were verified for small $n$ with the combstruct package.
with(combstruct);
gf :=
proc(n)
option remember;
local trees, evens;
trees := { T= Union(Z, Prod(Z, Set(T, card >=1 ))),
Z=Atom };
evens :=
proc(struct)
local rval;
if type(struct, function) then
if op(0, struct) = Prod then
return evens(op(2, struct))
else
rval := add(evens(op(q, struct)),
q=1..nops(struct));
if type(nops(struct), even) then
return 1+rval;
fi;
return rval;
fi;
fi;
return 1;
end;
add(u^evens(t), t in allstructs([T, trees, labeled], size=n));
end;
TGF := solve(T=z*exp(T), T);
q :=
proc(n)
option remember;
n!*coeftayl((TGF^2+z^2)/2/TGF/(1-TGF), z=0, n);
end;
