I think it just depends on how you want to use it. I will claim that sometimes the empty graph is best considered a tree and even a rooted tree but other times, neither. Even the one vertex tree is a little odd, it is the only tree with a degree zero vertex.
The Catalan numbers count many kinds of trees. In an Ordered Binary Tree each node may have up to two children left and/or right. If we let $C_n$ be the number of such with $n$ nodes then could exclude the empty tree and say
$C_1=1$
$C_{n+1}=C_n+C_n+\sum_i=1^{n-1}C_iC_{n-i}$
The first two terms for the case of only one child. But it is nicer to think of the left and right children as being themselves binary trees, both present, but perhaps one or both the empty tree.
$C_0=1$
$C_{n+1}=\sum_0^nC_iC_{n-i}$.
I think that the second approach is nicer. Particularly for the analogous situation with trinary trees.
A Full Ordered Binary Tree is as above except that a node may have either $0$ or $2$ children (thought of as nodes). There is a natural bijection between OBTs (including the empty tree) having $n$ nodes and FOBTs (not including the empty tree) having $n+1$ leaf nodes. In one direction assign each leaf node two children and in the other remove all the leaf nodes.
So here we interpret $C_n$ as the number of FOBTs with $n+1$ leaf nodes and do not bother to consider the empty tree as a FOBT.
Given a non-associative product, an expression $x_1\cdot x_2 \cdot x_k$ needs parentheses to be evaluated. We can use a FOBT with $k-1$ non-leaf nodes corresponding to the multiplications and $k$ leaves corresponding to the variables. Then $C_0$ counts the one vertex tree from the "product" $x_1.$ Now there seems no reason to count the empty tree. Of course we do like the empty product, but that is not especially relevant.
If we want to have a definition of rooted tree which does not specifically mention "the root" then we can say that a rooted tree is precisely a finite partial order $(S,\prec)$ such that
- for all $u \in S$ the set $\{x \mid x \preceq u\}$ is totally ordered by $\prec$
- for all $u,v \in S$ there is a common lower bound.
If we can get away with that, then the empty order is an order.