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 $n>1$.
Here is a more general version of my and YCor's argument:
Let $\Gamma$ be a group acting by measurable, volume-preserving transformations on a space $X$. Suppose that $\Gamma$ is amenable, in the sense that for each finite set $F$ and each $\epsilon>0$, there is a subset $S$ of $\Gamma$ such that $| FS| < (1+ \epsilon) |S|$ (Følner subsets).
Let $I$ and $J$ be two measurable subsets of $X$ such that can each be decomposed into finitely many pieces such that each piece of $I$ is a translate under an element of $\Gamma$ of a corresponding piece of $J$. Then the measure of $I$ equals the measure of $J$.
Proof: Let $F$ be the set of elements of $\Gamma$ that appear in the bijection between $I$ and $J$.
Let $f_I(x)$ be the number of $\gamma \in F S$ such that $\gamma(f) \in I$, and let $f_J(x)$ be the number of $\gamma \in S$ such that $\gamma(f) \in J$. Then $f_J(x) \leq f_I(x)$ because for each $\gamma$ with $\gamma(f) \in J$, we have chosen a $g\in F$ with $g \gamma(f) \in I$, and for distinct $\gamma(f)$, these give distinct elements of $I$ and hence distinct pairs $g \gamma$. (For distinct $\gamma$ that give the same $\gamma(f)$, we have the same $g$, and so $g \gamma$ remains distinct.)
So $\int f_J \leq \int f_I$. But exchanging the order of summation, $\int f_I = |FS| \mu(I)$ and $\int f_J = |S| \mu(J)$. Because so $(1+\epsilon) \mu(I) \leq \mu(J)$. Taking $\epsilon$ to $0$, $\mu(I) \leq \mu(J)$.
By symmetry, $\mu(I)=\mu(J)$.