Regular expressions and semi-linear sets
Your claim is correct for a one letter alphabet $\Sigma = \{a\}$. We prove by induction that $\psi(r)$ is linear for every regular expression without addition. When $r = a$, $\psi(r) = 1$. When $r = r_1 r_2$, by the induction hypothesis $\psi(r_1) = u_0 + \sum_i \mathbb{N} u_i$ and $\psi(r_2) = v_0 + \sum_j \mathbb{N} v_j$, and so $\psi(r) = u_0 + v_0 + \sum_i \mathbb{N} u_i + \sum_j \mathbb{N} v_j$. When $r = r_1^*$, $\psi(r)$ is a subset of $\mathbb{N}$ closed under addition. Let $g = \operatorname{gcd}(\psi(r))$. Then $\psi(r)/g$ is a numerical semigroup, and so finitely generated, say $\psi(r)/g = \sum_i \mathbb{N} u_i$. So $\psi(r) = \sum_i \mathbb{N} gu_i$.

This argument doesn't quite work when $|\Sigma| \geq 2$, although there are some notions of GCD for multidimensional vectors, see for example this interesting presentation by Vadim Ponomarenko. Perhaps the argument above can be pushed through, or perhaps a different argument, which uses the fact that $\psi(r_1)$ is linear, exists. It's also possible that the claim is wrong in this case.