Late edit: having now read through the OP comments, I can see that my proof is essentially a carbon copy of @darij.grinberg's approach (although my derivation was independent). I'm okay to delete this answer once/if darij chooses to post theirs.
Pick a canonical ordering of unlabelled rooted trees, say, with lexicographical comparison of tuples $(n, T_1, \ldots, T_k)$, where $n$ is the number of vertices, $T_1, \ldots, T_k$ is the non-descending sequence of children subtrees. Throughout $T, T_1, T_2$ are unlabelled rooted trees in canonical form (that is, subtrees of any vertex are ordered as above).
Let $A_n$ be the set of pairs $(T, v)$ with $|T| = n$, $v$ is a non-root vertex of $T$. Also, let $B_n$ be the set of tuples $(T_1, T_2, k, u)$ such that $|T_1| + k|T_2| = n$, $u$ is a vertex of $T_2$. Observe that $|A_n|$ and $|B_n|$ are LHS and RHS of the recurrence in OP, more readily seen by rewriting $(n - 1)t_n = \sum_{m = 1}^{n - 1}mt_m \sum_{0 < km < n} t_{n - km}$.
The bijection between $A_n$ and $B_n$ is as follows:
- we associate $(T_1, T_2, k, u)$ to $(T, v)$ by:
- $T_2$ = the subtree of $T$ containing $v$,
- $u$ = the respective vertex of $T_2$,
- $k$ = the number of children subtrees of $T$ isomorphic to $T_2$ not later than the copy containing $v$ in the ordered sequence of children subtrees,
- $T_1$ = the result of removing the $k$ children subtrees from $T$.
- we associate $(T, v)$ to $(T_1, T_2, k, u)$ by inserting $k$ copies of $T_2$ as children subtrees of the root of $T_1$ (and naming the result $T$), and picking the respective vertex $u$ in the $k$-th copy as $v$.