Let's extend the language of ZF with a new relation-symbol $\le$ and add axioms asserting that it is a total ordering:
$\forall_x (x \le x)$
$\forall_x \forall_y (x \le y \vee y \leq x)$
$\forall_x \forall_y ((x \le y \land y \le x) \rightarrow x = y)$
$\forall_x\forall_y\forall_z((x \le y \land y \le z) \rightarrow x \le z)$
Suppose there is a strongly inaccessible cardinal $\kappa$, so $V_\kappa$ is a model of ZF. Then I believe there is also a model of the extended theory, since we can totally order $V_\kappa$ in some arbitrary way and use that as our interpretation of $\leq$. A technicality is that Separation now has additional instances asserting the existence of new sets like $\{ (x,y) \in \mathbb{N}^2 : x \leq y \}$. But all these sets were already in $V_\kappa$ so $V_\kappa$ is also a model of the extended theory which is therefore a conservative extension of ZF. Stop me there if any of the above is wrong.
The general question motivating what follows is what can we say to relate $\leq$ and $\in$ and keep the theory consistent?
For example I claim we can add the following axioms as well:
$A \equiv \forall_x \forall_y ( x \leq y \leftrightarrow (x \setminus y) \leq (y \setminus x))$
$B \equiv \forall_x \forall_y ((\exists_z(z \in y \land \forall_w(w \in x \rightarrow w \lt z))) \rightarrow x \lt y)$
$C \equiv \forall_x \forall_y (\text{rank}(x) \subset \text{rank}(y) \rightarrow x \lt y)$
So one sufficient condition for $x \lt y$ is $x \subset y$ (implied by $A$), another is that $y$ contains an element larger than every element in $x$ ($B$), and another is that $x$ has a smaller rank than $y$ ($C$). This suffices to define $\leq$ uniquely on $V_\omega$ (the hereditarily finite sets are ordered as Ackermann numbers) and by transfinite induction on rank we can give an interpretation of $\leq$ under which $A$ and $B$ and $C$ hold for all sets. For each rank $V_{\alpha+1}\setminus V_\alpha$, we already know how to compare its elements with elements of earlier ranks by $C$ (they are always greater), so we just need to be able to compare them with each other. Now $B$ defines a partial ordering on disjoint sets with the same rank, whose definition extends to the entire rank using $A$. Finally choose a total ordering on $V_{\alpha+1}\setminus V_\alpha$ containing that partial ordering to define $\leq$ in the model: the set $P((V_{\alpha+1} \setminus V_\alpha)^2)$ contains all relations on $V_{\alpha+1}\setminus V_\alpha$, so this can be accomplished using Specification and Choice.
Is that a valid argument that this theory is consistent if ZF is?