You can get a general answer avoiding the discussion on the choice of the generating set or of the metric on the larger space: which finitely generated (f.g.) groups have a bilipschitz embedding into a Euclidean space?

The answer is: virtually abelian f.g. groups (i.e. having a f.g. abelian group of finite index).

Indeed such groups admit proper cocompact actions on $\mathbf{R}^n$ for some $n$, giving rise to a quasi-isometric embedding which can easily be deformed to a bilipschitz embedding (alternatively, if you like to stay with an action, you can add some extra dimensions to make the action [whose kernel is finite] faithful but you lose cocompactness).

Conversely, if a group $\Gamma$ admits a bilipschitz embedding (coarse embedding --aka uniform embedding-- would be enough) into a Euclidean space, by a growth argument, the group has polynomial growth. So it is virtually nilpotent, by Gromov's polynomial growth Theorem. Then a result of general result of Pauls (2001) shows that a virtually nilpotent f.g. group admits a bilipschitz embedding into a CAT(0) space (or a uniformly convex Banach space). However, I guess that this follows, in this special case (Euclidean target) from Pansu's 1989 Annals paper about metric differentiability. The argument consists of observing that a bilipschitz map $\Gamma\to\mathbf{R}^n$ induces a bilipschitz embedding from the asymptotic cone (which is a simply connected nilpotent group $G$ with a Carnot-Caratheodory metric, by an older result of Pansu) into $\mathbf{R}^n$, to show that generically it is metrically differentiable and the differential is generically an injective homomorphism; in particular $G$ is abelian, and this holds iff the original discrete group is virtually abelian.

To go back to the original question: if you stick to isometric embeddings, what you get as a corollary is that if $\Gamma$ is a group with a finite generating set $S$, and $E$ is a finite-dimensional Banach space, then if the Cayley graph of $\Gamma$ w.r.t. $S$ embeds isometrically into $E$ then $\Gamma$ is virtually abelian (which is an extremely restrictive condition). The converse, however, depends on the specific choice of $S$ and the norm on $E$, as pointed out in other answers.