Suppose $\Gamma$ is a discrete, cocompact Abelian subgroup of the Lie group $\mathrm{E}(n) = \mathbb R^n \rtimes \mathrm{O}(n)$ of Euclidean isometries, where $$(b,A)(b',A') = (b+Ab', AA') \quad \forall \: (b,A), (b',A') \in \mathrm{E}(n)$$ I would like to prove:
If $\Gamma$ acts freely on $\mathbb R^n$, then $\Gamma$ is a lattice in $\mathbb R^n$. In particular, for every $(b,A) \in \Gamma$, we have $A = \mathrm{Id}_n$.
Because $\Gamma$ acts freely on $\mathbb R^n$, if $(b,A) \in \Gamma$, then the equation $b + Ax = x$ has no solution; in particular, $b$ is not in the image of $\mathrm{Id}_n - A$. So $A$ must have an eigenvalue of $1$. But this is about as far as I've been able to get. Any thoughts?
EDIT: As was pointed out to me, cocompactness is a necessary assumption. I found a good exposition of this result in Louis Auslander's AMS article "An Account of the Theory of Crystallographic Groups".
Not an answer
(Or, more accurately, an answer to the question before you changed it...)
Skipping the co-compactness that you added later, pick $b = \pmatrix{0\\0\\1}, A = \pmatrix{0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1}$ and you've got the screw translation mentioned in the comments. The slab $\Bbb R^2 \times [0, 1]$ is a fundamental domain. $A$ does in fact have $1$ as an eigenvalue, but that's not enough to get you the "translation" conclusion that you want.
In short: co-compactness appears to be essential. *