Here’s a proof by induction.
Let $w(n) = \sum\limits_{G\in C_n}(-1)^{e(G)}$, where $e(G)$ is the number of edges in $G$. I will assume that the vertex set of $G\in C_n$ is $[n] = [1,n]\cap\mathbb{Z}^+$. Suppose that $w(k) = (-1)^{k-1}(k-1)!$ for $1\le k \le n$.
Let $H$ be a labeled partition of $[n]$. Suppose that $H$ has components (parts) $H_1,\dots,H_k$. Denote $v_i = |H_i|$. Let $P(H)$ be the set of permutations whose cycles are exactly $H_1,\dots,H_k$ (in any internal order). Clearly $$\vert P(H)\vert = \prod_{i=1}^k(v_i-1)!,$$ and if $G_n$ is the set of all labeled partitions of $[n]$, $$\sum_{H\in G_n}\vert P(H)\vert = n!.$$
Let $C_{n+1}(H)$ be the set of $G\in C_{n+1}$ that decompose into connected components $H$ after deleting vertex $n+1$. Consider a component $H_i$ of $H$. If $G\in C_{n+1}(H)$, vertex $n+1$ must be adjacent in $G$ to at least one vertex of $H_i$. For $j=1,\dots,v_i$ there are $\dbinom{v_i}{j}$ ways to attach vertex $n+1$ to $j$ vertices of $H_i$. It follows that
$$\begin{align*}
\sum_{G\in C_{n+1}(H)}(-1)^{e(G)} &= \prod_{i=1}^k\sum_{j=1}^{v_i}\binom{v_i}{j}(-1)^jw(v_i)\\
&= \prod_{i=1}^k(-1)^{v_i-1}(v_i-1)!\sum_{j=1}^{v_i}\binom{v_i}{j}(-1)^j\\
&= \prod_{i=1}^k(-1)^{v_i-1}(v_i-1)!(-1)\\
&= \prod_{i=1}^k(-1)^{v_i}(v_i-1)!\\
&= (-1)^n\prod_{i=1}^k(v_i-1)!\\
&= (-1)^n\vert P(H)\vert
\end{align*}$$
and hence that
$$\begin{align*}
w(n+1) &= \sum_{H\in G_n}\quad\sum_{G\in C_{n+1}(H)}(-1)^{e(G)}\\
&= \sum_{H\in G_n}(-1)^n \vert P(H)\vert\\
&= (-1)^n\sum_{H\in G_n}\vert P(H)\vert\\
&= (-1)^n n!.
\end{align*}$$
Marko’s comprehensive answer is a bit overwhelming if you’ve not encountered that approach; let me see if I can offer one closer to what you’ve seen before.
Once you have a root, you can hang three trees from it: one is its left subtree, another is its centre subtree, and the third is its right subtree. (These are ordered trees, so left, centre, and right are meaningful here.) Any of these subtrees can of course be empty. Suppose that the left subtree has $i$ nodes, the centre one has $j$ nodes, and the right one has $k$ nodes, and you’re building a tree with $n$ nodes; then $i+j+k=n-1$, the root being the $n$-th node. This is the decomposition requested in part (B), and it gives you the recurrence
$$t_n=\sum_{\large{{0\le i,j,k}\atop{i+j+k=n-1}}}t_it_jt_k=\sum_{i=0}^{n-1}t_i\sum_{j=0}^{n-1-i}t_jt_{n-1-i-j}\;.\tag{1}$$
The problem now is to convert this into an equation that $T(x)$ must satisfy.
The last summation in $(1)$ is clearly one term of the convolution of the sequence $\langle t_n:n\in\Bbb N\rangle$ with itself; specifically, it’s the coefficient of $x^{n-1-i}$ in $\big(T(x)\big)^2$. The same reasoning shows that the entire right-hand side is the coefficient of $x^{n-1}$ in $\big(T(x)\big)^3$, so it’s the coefficient of $x^n$ in $x\big(T(x)\big)^n$. $(1)$ says that it’s also the coefficient of $x^n$ in $T(x)$, so to a first approximation we have $T(x)=x\big(T(x)\big)^3$. A little care is needed here, however, because $x\big(T(x)\big)^3$ has a $0$ constant term, and we know that $t_0=1$; thus, the correct relationship is
$$T(x)=1+x\big(T(x)\big)^3\;.$$
Note that with a bit more experience you won’t have to go through to convolution steps: you’ll be able to recognize that summing over $m$ indices whose sum is constant is basically taking an $m$-th power of the generating function.
Best Answer
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
$$ T(x)=1+A(x)-\frac12\left(A(x)^2-A(x^2)\right)\;. $$