[The following is not quite an answer, but it refutes a natural
generalization suggested in the Comments, and is too long to be
a comment itself.]

Counterexample in ${\bf R}^N \times {\bf R}$ for some $N>2$:
any lattice hexagon $H$ with angles
$90^\circ$, $90^\circ$, $135^\circ$, $135^\circ$, $135^\circ$, $135^\circ$
in that order. (That is, $H$ is obtained from a lattice rectangle by
truncating two adjacent vertices by two isosceles right lattice triangles,
not necessarily congruent. Alternatively, remove two congruent
lattice triangles, not necessarily isoceles, related by a 90-degree rotation.)
Then $H$ does not tile the plane, but
four copies do tile a (non-simply-connected) polyomino. But it was
recently shown that *any* polyomino in some ${\bf Z}^n$
(which need not be simply connected, or even connected at all!)
tiles ${\bf Z}^d$ for some $d$:

Vytautas Gruslys, Imre Leader, and Ta Sheng Tan: Tiling with arbitrary tiles.
*Proc. London Math. Soc.* (2016) **112** (6): 1019-1039.
https://doi.org/10.1112/plms/pdw017 $\cong$ http://arxiv.org/abs/1505.03697

(I learned about this from **Francisco Santos**'s accepted answer to
**Timothy Chow**'s
Mathoverflow question 49915, which the MO algorithm helpfully put at
the top of its list of questions "Related" to this one.)

2more comments