Why must an isotoxal polyhedron or tiling be isogonal or isohedral? The Wikipedia page "Isotoxal Figure" said that an edge-transitive polyhedron or tiling must be vertex-transitive or face-transitive. I deleted that because it is false in general, as in the following example. Under what further conditions is it true, and how is it proven?
Consider an irregular spherical dihedron, made of a small rhombus in the Northern hemisphere, and a large rhombus on the rest of the sphere, including the Equator and the Southern hemisphere. The dihedron's symmetries are : two perpendicular reflections (across the rhombus' diagonals), the resulting 180° rotation (around the rhombus' center), and the identity. These four symmetries map any edge to any other, so it is edge-transitive. But a vertex only gets mapped to itself and the opposite vertex, not the adjacent two, so it is not vertex-transitive. And, of course, the two rhombi have different areas, so the dihedron is not face-transitive either. Thus this spherical polyhedron is isotoxal, but neither isogonal nor isohedral.
I just realized that there is a distinction between combinatorial (abstract polytope) symmetries, and geometric symmetries. This may be a part of the answer. The above dihedron is regular as an abstract polyhedron (equivalent to the square dihedron).
EDIT1 :
Here are pictures of some more examples of only-edge-transitive polyhedra. The first group of spheres is dihedra. The second group is similar to hosohedra, except that the lunes' edges are split in two, so each face has four edges instead of two. (Green lines are edges, and black dotted lines are reflection symmetries.)

Notice that all of these polyhedra have some vertices of valence 2.
Another example can be found in the rhombic dodecahedral honeycomb. Take one layer of dodecahedra (and look at only the top four faces of each).

I have stretched everything by different factors along the three axes, so the rhombi have two different shapes. Notice that this is not a plane tiling (which probably must never be only-edge-transitive*), but a 3D apeirohedron which is only-edge-transitive.
*A plane tiling may be only-edge-transitive if it is colored to break the symmetry. An example of this (probably the only example; see Jaap's comment) is the above apeirohedron with a vertical scale factor of 0, and the colors maintained despite equal-shaped rhombi.
EDIT2:
I will provide an equation for the rhombic apeirohedron. Define the triangular wave as $t(x) = 1 - \vert x \vert,$ if $\vert x \vert \le 2,$ and $t(x+4) = t(x),$ for all $x$. Then the surface is $$z = c \frac{t \left(\frac xa - \frac yb \right) + t \left(\frac xa + \frac yb \right)}{2}.$$
EDIT3:
Another example is a hyperbolic tiling with a quadrilateral fundamental domain. The quad's angles are $ \left( \frac{ \pi}{p}, \frac{ \pi}{p}, \frac{ \pi}{q}, \frac{ \pi}{q} \right) $ (with adjacent angles, not opposite angles, congruent), and the sum should be $ \frac{2 \pi}{p} + \frac{2 \pi}{q} < 2 \pi $. The tiling is made by taking a diagonal of the quadrilateral (the "seed" edge for the tiling), reflecting it across the quad's edges, and repeatedly reflecting the newly generated edges of the tiling.
This hyperbolic tiling has $(2p)$-gon and $(2q)$-gon faces, and vertices of valence $2p$ and $2q$. It is only-edge-transitive, both geometrically and combinatorially (which brings me to my final example).
A finite abstract polyhedron, without any geometry, can be only-edge-transitive. I've modeled it as a spherical polyhedron, but the edges are not geodesics. It has $3$ square faces, $6$ digon faces, $3$ valence-4 vertices, $2$ valence-6 vertices, and $12$ edges.

More of these can be gotten from the "almost-hosohedra" shown above, by opening up the edges into digons.
 A: I will try to answer my own question, but I might accept another answer.
Any isotoxal polyhedron must have at most two types of faces or vertices, because there is only one type of edge, and an edge connects exactly two vertices and two faces. If it is not isogonal, it cannot have one vertex type, so it must have exactly two; call them $P$ and $Q$. Likewise, if it is not isohedral, it cannot have one face type, so it must have exactly two; call them $A$ and $B$.

Any symmetry of the polyhedron must send vertices to vertices, and faces to faces. For a symmetry which sends an edge to an adjacent edge (sharing a vertex and a face), there are four possibilities:

The first three would send $B$ to $A$, or $Q$ to $P$, making them equivalent. Only the fourth symmetry keeps two distinct face and vertex types. This symmetry maps an angle in $A$ (at $Q$) to the second one over, so every other angle in $A$ must be congruent, and every other vertex of $A$ must be equivalent. If $A$ had an odd number of sides, this would make $Q$ equivalent to $P$, so $A$ must have an even number of sides; call this $2 n_A$. By similar reasoning, vertex $P$ must have two angle types, and an even number of faces around it, alternating type $A$ and type $B$; call this number $2 n_P$. Define $2 n_B$ and $2 n_Q$ analogously.
As $A$ has only two angle types, corresponding to the vertex types, call them $\alpha_P$ and $\alpha_Q$. So far, we have not assumed any particular geometry (Euclidean, spherical, or hyperbolic). For now, assume that it is a non-self-intersecting Euclidean polyhedron. For any Euclidean $k$-gon, the sum of the vertex angles is $(k - 2) \pi$ [radians], so $$n_A \alpha_P + n_A \alpha_Q = (2 n_A - 2) \pi$$ $$\frac{\alpha_P + \alpha_Q}{2} = \frac{2 n_A - 2}{2 n_A} \pi$$ or, the average of $\alpha_P$ and $\alpha_Q$ is equal to the angle of a Regular $(2 n_A)$-gon. Likewise, for angles $\beta$ in face type $B$, $$\frac{\beta_P + \beta_Q}{2} = \frac{2 n_B - 2}{2 n_B} \pi$$ We can re-arrange these equations to get $$\alpha_Q = \frac{n_A - 1}{n_A} 2 \pi - \alpha_P,$$ $$\beta_Q = \frac{n_B - 1}{n_B} 2 \pi - \beta_P$$
The sum of the angles around vertex $P$ is $n_P \alpha_P + n_P \beta_P$, so the angle defect is $$\delta_P = 2 \pi - n_P (\alpha_P + \beta_P)$$ and $$\delta_Q = 2 \pi - n_Q \alpha_Q - n_Q \beta_Q$$ $$= 2 \pi - n_Q \frac{n_A - 1}{n_A} 2 \pi + n_Q \alpha_P - n_Q \frac{n_B - 1}{n_B} 2 \pi + n_Q \beta_P$$ $$= \left(1 - 2 n_Q + \frac{n_Q}{n_A} + \frac{n_Q}{n_B} \right) 2 \pi + n_Q (\alpha_P + \beta_P)$$
Then the average angle defect is $$\frac{\delta_P + \delta_Q}{2} = \frac{1}{2} \left( \left(2 - 2 n_Q + \frac{n_Q}{n_A} + \frac{n_Q}{n_B} \right) 2\pi + (n_Q - n_P) (\alpha_P + \beta_P) \right)$$ $$= n_Q \left( \frac{2}{n_Q} + \frac{1}{n_A} + \frac{1}{n_B} - 2 \right) \pi + (n_Q - n_P) \frac{\alpha_P + \beta_P}{2}$$
Without loss of generality, we can assume that $n_P \geq n_Q$, so $n_Q - n_P \leq 0$. The angles are all positive, so $(n_Q - n_P) \frac{\alpha_P + \beta_P}{2} \leq 0$. Also, because the polyhedron is Euclidean and non-self-intersecting, $2 n \geq 3$, for each of $n_A$, $n_B$, $n_P$, and $n_Q$. (A spherical polyhedron may have $2 n = 2$, and a self-intersecting star polygon has fractional $n$.) So $$n \geq 2$$ $$\frac{1}{n} \leq \frac{1}{2}$$ $$\frac{2}{n_Q} \leq 1, \frac{1}{n_A} \leq \frac{1}{2}, \frac{1}{n_B} \leq \frac{1}{2}$$ $$\frac{2}{n_Q} + \frac{1}{n_A} + \frac{1}{n_B} \leq 1 + \frac{1}{2} + \frac{1}{2} = 2$$ $$\frac{2}{n_Q} + \frac{1}{n_A} + \frac{1}{n_B} - 2 \leq 0$$ $$n_Q \left(\frac{2}{n_Q} + \frac{1}{n_A} + \frac{1}{n_B} - 2 \right) \pi \leq 0$$ $$n_Q \left(\frac{2}{n_Q} + \frac{1}{n_A} + \frac{1}{n_B} - 2 \right) \pi + (n_Q - n_P) \frac{\alpha_P + \beta_P}{2} \leq 0$$ $$\frac{\delta_P + \delta_Q}{2} \leq 0$$
(This inequality becomes an equality only if all $n = 2$; this corresponds to the rhombic apeirohedron.) Any Euclidean polyhedron with spherical topology has total angle defect $\Sigma \delta = 4 \pi > 0$, so this polyhedron cannot have spherical topology. It may have the topology of a torus, or a double torus, triple torus, etc., or it may be infinite.
If it is a spherical tiling instead of a Euclidean polyhedron, the same approach can be modified:
For any spherical $k$-sided polygon, the sum of the angles is, not equal to, but greater than $(k - 2) \pi$. So $$n_A \alpha_P + n_A \alpha_Q > (2 n_A - 2) \pi$$ $$\alpha_Q > \frac{n_A - 1}{n_A} 2 \pi - \alpha_P$$ $$\beta_Q > \frac{n_B - 1}{n_B} 2 \pi - \beta_P$$ Also, because it is a 2D tiling, the angle defect is 0; $$2 \pi = n_P \alpha_P + n_P \beta_P$$ $$2 \pi = n_Q \alpha_Q + n_Q \beta_Q > n_Q \left( \frac{n_A - 1}{n_A} 2 \pi - \alpha_P + \frac{n_B - 1}{n_B} 2 \pi - \beta_P \right)$$ $$ \frac{2 \pi}{n_Q} > \left(1 - \frac{1}{n_A} + 1 - \frac{1}{n_B} \right) 2 \pi - (\alpha_P + \beta_P) = \left(2 - \frac{1}{n_A} - \frac{1}{n_B} \right) 2 \pi - \frac{2 \pi}{n_P}$$ $$ \frac{1}{n_Q} + \frac{1}{n_A} + \frac{1}{n_B} + \frac{1}{n_P} > 2$$
As before, if each $n \geq 2$, then $\frac{1}{n} \leq \frac{1}{2}$, and $\Sigma \frac{1}{n} \leq 2$, which contradicts the above inequality. So there must be some $n = 1$. That means there is either a vertex of valence $2 n = 2$, or a lune with $2$ edges. But if there is a lune, which has edge length $\pi$, every edge must have length $\pi$, because of edge-transitivity. Then every face must be a lune, and there are only 2 vertices (call them North and South poles), and the tiling has a symmetry of reflection across the equator, which makes the two vertices equivalent (unless the edges are directed, or the vertices are colored, to maintain asymmetry). So an (un-colored) only-edge-transitive spherical tiling must have a vertex of valence $2$.
