Consider the following theorem:
Whenever a rectangle is tiled by rectangles each of which has at least one integer side, then the tiled rectangle has at least one integer side.
There is a paper by Stan Wagon with fourteen proofs of this result. If we replace integer by rational, then it is possible to prove this result using tensor products.
Sketch of a proof: ($\otimes = \otimes_\mathbb{Q}$) Let $a$ and $b$ the sides of the big rectangle and let $(a_i)$, $(b_i)$ the sides of the small rectangles. We have $a\otimes b=\sum a_i\otimes b_i$, using that each rectangle has a rational side we can rewrite this sum as $1\otimes c + d\otimes 1$. Let $f:\mathbb{R}\to\mathbb{R}$ be a $\mathbb{Q}$-linear function such that $\ker f =\mathbb{Q}$ (it is possible to construct it using a $\mathbb{Q}$-basis of $\mathbb{R}$). Now, define $\phi:\mathbb{R}\otimes \mathbb{R} \to \mathbb{R}$ as $\phi(u\otimes v)=f(u)f(v)$. We have $$ f(a)f(b)=\phi(a\otimes b)=\phi(1\otimes c + d\otimes 1)=0, $$ so that $a\in \mathbb{Q}$ or $b\in \mathbb{Q}$.
Question: Is it possible to write a similar proof for the original question?
Comment: As $\mathbb{R}\otimes_\mathbb{Z} \mathbb{R} = \mathbb{R}\otimes_\mathbb{Q} \mathbb{R}$ I would think the answer is no, but maybe we can overcome this problem using different $\mathbb{Z}$-modules.
There is a version which handles the integer case, see A. Shen, An unfair game. Colorings and coverings. Tilings and Polyhedra revisited. Math. Intelligencer, 19 (1997), no.4, p. 48--50. Shen quotes a letter from Deligne with the following proof. (Everything below is taken from this paper.)
Consider the direct sum $\mathbb Z^{(\mathbb R/\mathbb Z)}$ of copies of $\mathbb Z$, indexed by $\mathbb R/\mathbb Z$. For an element $x\in\mathbb R$, denote by $\delta(x)$ the "standard basis vector" in $\mathbb Z^{(\mathbb R/\mathbb Z)}$ corresponding to the class of $x$ in $\mathbb R/\mathbb Z$.
Now given a rectangle whose sides are parallel to the coordinate axes, say $A=[x_0, x_1] \times [y_0, y_1]$, consider the element $c(A) := (\delta(x_1)-\delta(x_0))\otimes (\delta(y_1)-\delta(y_0))$ in the tensor product $\mathbb Z^{(\mathbb R/\mathbb Z)}\otimes\mathbb Z^{(\mathbb R/\mathbb Z)}$. Then $c(A)$ behaves additively when $A$ is partitioned into smaller rectangles.
Clearly $c(A)=0$ whenever one of the two side lengths is integral. It is not hard to check that the converse is true: If $c(A)=0$, then one of the two side lengths of $A$ is integral. (As Deligne insists, one does not need the axiom of choice here; similarly in the reasoning given in the question it is not actually necessary to use a $\mathbb Q$-basis of $\mathbb R$ - instead one can work with a suitable finite-dimensional subvector space of $\mathbb R$.)
Note that this method of proof allows replacing the integers by any subgroup of $\mathbb R$, and that even two different subgroups for the two axes can be used!