Does foundation/regularity have any categorical/structural consequences, in ZF?

Yes, the axiom of foundation has structuralist consequences.

Let $\phi$ be the assertion, "if every well-founded set is well-orderable, then every set is well-orderable."

This statement, I claim, can be made in a (technically) structuralist manner, since it is a statement about all nodes in a certain kind of digraph, namely, the graphs that are stratified by a well-ordered sequence of levels, such that every new level has a unique node for every subset of the nodes on the previous levels. And then assert that every node in such a graph has another node that amounts to a well-order of the set pointing at that node. That is, just make a structuralist description of the $V_\alpha$ hierarchy, and assert that if all the sets arising in it are well-orderable, then every set is well-orderable.

Certainly ZF proves the statement $\phi$, since ZF proves that every set arises in the well-founded $V_\alpha$ hierarchy.

But ZF-AF does not prove the statement, since, I believe, it is consistent with ZF-FA that the well-founded part of the universe satisfies AC, even if there is a non-well-orderable set of Quine atoms in the non-well-founded part of the universe. One can build such a model, I believe, as a symmetric extension of a model of ZFC, by adding a collection of urelements, considered as Quine atoms.

The following provides an "engine" for generating many structural statements provable in ZF but not in ZF without foundation:

Theorem. Let $S$ be any structural equivalent of the axiom of choice within ZF without foundation (e.g., Zorn's Lemma or Tychonoff's compactness theorem). Then the following structural statement $R_{S}$ (in honor of Herman and Jean Rubin) is provable in ZF; but NOT in ZF without foundation:

$R_{S}$ := "$S$ holds iff every linearly orderable set is well-orderdable".

Explanation. The nontrivial direction ($\leftarrow$) of $R_{S}$ follows from the theorem below, and the observation that the powerset of every well-orderable set $S$ is linearly orderable (by deeming $X$ to be less than $Y$, where $X$ and $Y$ are subsets of $S$, and $S$ well-ordered by $<$, if the $<$-first element in the symmetric difference of $X$ and $Y$ is in $Y$. The proof is readily implementable in Zermelo set theory, i.e., ZF without the foundation axiom and without the replacement cheme, but with the separation scheme).

Theorem (H. Rubin & J. Rubin). ZF proves that if the powerset of every well-orderable set is well-orderable, then the axiom of choice holds.

The above theorem was first published the 1963 book Equivalents of the Axiom of Choice, by Herman and Jean Rubin. A proof can also be found in these lecture notes by Andrés E. Caicedo (see Theorem 13). Later, in this 1973 paper of Felgner & Jech, a model of ZF with atoms was built in which the powerset of every ordinal is well-orderable, but the axiom of choice fails; this proof also appears in chapter 9 of the book Axiom of Choice by Jech.