Could we use combinatorics and generating functions to calculate the proportion of n-node trees whose root has only one or two subtrees? Here is what I tried:
The combinatorial construction for the trees with a single or two subtrees (at the root) is:
$ R = Z \cdot Z \cdot SEQ(G) + Z \cdot SEQ(G) \cdot SEQ(G) $
Thus we have:
$ R(z) = z^2 \frac{1}{1-G(z)} + z \frac{1}{[1-G(z)]^2} $
And we know that:
$ G(z) = z \cdot \frac{1+\sqrt{1-4z}}{2}$
I cannot continue since the expression became complicated.
This is possible, essentially by following the technique you suggested. Let $G(z)$ be the generating function for the number of $n$-node unlabelled plane trees. It's well known (see http://en.wikipedia.org/wiki/Catalan_number) that this number is just the Catalan number $C_{n-1}$, so we have that
$$G(z) = z\cdot\dfrac{1-\sqrt{1-4z}}{2z} = \dfrac{1-\sqrt{1-4z}}{2}$$
From this, we can see that the generating function for $n$-node plane trees where the root has one child is given by $H_{1}(z) = zG(z)$, and the generating function for $n$-node plane trees where the root has two ordered children is given by $H_{2}(z) = zG(z)^2$. Summing these two expressions, the generating function for the value you wish to count is given by
$$H(z) = H_1(z) + H_2(z) = z(1 - z - \sqrt{1-4z})$$
By expanding out $\sqrt{1-4z}$ according to the binomial theorem (similarly as in the derivation of Catalan numbers), we have that $\sqrt{1-4z} = 1 - 2\sum_{n=1}^{\infty}\frac{1}{n}\binom{2(n-1)}{n-1}z^{n}$. Substituting this in, we find that
$$H(z) = z^2 + 2\sum_{n=3}^{\infty}\binom{2(n-2)}{n-2}\dfrac{z^{n}}{n-1}$$
It follows, that $h_{n}$, the number of plane trees with $n$ nodes and the root having one or two children equals (for $n>2$)
$$h_{n} = \dfrac{2}{n-1}\binom{2(n-2)}{n-2} = 2C_{n-2}$$
Since the total number of plane trees on $n$ nodes is $C_{n-1}$, trees with one or two children at the root form a proportion of
$$\dfrac{h_{n}}{C_{n-1}} = \dfrac{2C_{n-2}}{C_{n-1}} = \dfrac{n}{2n-3}$$
of the total number of such trees.