No. Let $A$ be the free abelian group generated by the $r_a$s. We can view this as a lattice in the real vector space $A \otimes \mathbb R$. Let $n$ be the rank of this group / the dimension of this vector space.


There is a natural evaluation map $f: A \otimes \mathbb R \to \mathbb R$ from this vector space to $\mathbb R$. Applying the assumed bijection between $[2,3] \cup [0,1] $ and $[0,1]$, which is obtained by at each point by subtracting one of the $r_a$s and hence preserved the property of lying in $A$, we obtain a bijection between $A \cap f^{-1} ( [2,3] \cup [0,1])$ and  $ A \cap f^{-1} ([0,1])$.

In the case $n=1$, this is already impossible as soon as the lattice intersects $[2,3]$, which we can guarantee if necessary by adding additional $r_a$s.

Otherwise, this bijection involves moving in the lattice a distance at most $1$. Hence for $B_R$ a ball of radius $R$, we obtain an injection $B_R \cap A \cap f^{-1} ([2,3] \cup [0,1]) \to  B_{R+1} \cap  A \cap f^{-1} ([0,1])$.

We now simply apply lattice point counting to check that the number of lattice points in these sets are asymptotic to the volumes of  $B_R \cap f^{-1} ([2,3] \cup [0,1]) $ and $  B_{R+1}  \cap f^{-1} ([0,1])$ respectively, which are asymptotic to $2 C R^{n-1}$ and $C R^{n-1}$ respectively, contradicting the claimed existence of an injection.

The only subtlety in the lattice point counting is whether the images of the lattice points under $f$ are equidistributed. However, by Weyl equidistribution, it suffices that $A$ is a dense subset of $\mathbb R$, which is automatic as $n>1$.