Tiling a rectangle with a hint of magic

It is not at all obvious to me that there is any deep principle at work in the double-integral proof. In my mind, the double-integral proof is really the same as the checkerboard proof. You're just trying to come up with a translation-invariant function on rectangles that is (a) additive and (b) zero if and only if the rectangle has the property of interest. All that matters is that you pick a function that cancels itself out in the same way as the checkerboard pattern does. The choice of the sine function for this purpose is amusing but not deep.

If you're looking for deeper principles then I would recommend Rick Kenyon's paper "A note on tiling with integer-sided rectangles," J. Combin. Theory Ser. A 74 (1996), no. 2, 321-332. Using this problem as an example, Kenyon demonstrates the concept of the Conway-Lagarias tiling group, a powerful tool for studying tiling problems.

There is an exercise in "Modern Graph Theory" by Bollobas section II.4 pg 63 that is essentially the same argument, but eliminates the trigonometric functions. For each rectangle $U = [x_1, x_2] \times [y_1, y_2]$ let $\psi(U) = (x_2 - x_1) \otimes (y_2 - y_1)$ in $\mathbb{Z}(\mathbb{R}/\mathbb{Z}) \otimes \mathbb{Z}(\mathbb{R}/\mathbb{Z})$ (viewed as a $\mathbb{Z}$ module). Then $\sum_{U} \psi(U) = 0$ so the original rectangle must have an integer side.

[Edited after Reid's comment. Here $\mathbb{Z}(\mathbb{R}/\mathbb{Z})$ is the free $\mathbb{Z}$ module with basis $\mathbb{R}/\mathbb{Z}$]

There is an unpublished note on this theorem from 1987 by Edgser W. Dijkstra, On a Problem transmitted by Doug McIlroy, see also this summary. He explains how the proof that uses complex double integrals (number 1 in Wagon's paper) naturally follows from an initial choice of formalizing the tiling property, and in particular, how the complex numbers come up. In the end he speculates about the general applicability of such an approach.

As an aside, there is also a proof with contour integrals in a much later note of Dijkstra: Ulrich Berger’s argument rephrased. ("Ulrich Berger’s argument" is essentially the sweep-line proof, number 12 in Wagon's paper, see EWD1023, Ulrich Berger's solution to the rectangle problem.)