Let $A(x)=1+xB(x) \in \mathbb{C}[[x]]$ be a power series. A standard calculation shows that $$A(x) = \frac{\prod (1-\alpha_i x)}{\prod (1-\beta_j x)} \leftrightarrow x \frac{A'(x)}{A(x)} = \sum_{n \ge 1} (\sum a_i^n - \sum b_j^n)x^n.$$ If we write $A(x)$ as an infinite product $$A(x) = \prod_{n \ge 1} (1-x^n)^{-a_n}$$ for some $a_n \in \mathbb{N}$ (this is always possible, in a unique way), we have $$x \frac{A'(x)}{A(x)} = \sum_{n \ge 1} (\sum_{d \mid n} a_d d)x^n.$$
Thus, what we really need to understand is $\sum_{d \mid n} a_d d$ rather than $a_n$. In our case, $a_{2n} = SRMI_q(2n), a_{2n-1} = 0$. I will focus on the even $q$ case, for simplicity. Let $n$ be a positive integer. We have, by changing the order of summation,
$$\sum_{d \mid n} a_d d =\sum_{2d \mid n} a_{2d} 2d=\sum_{m \mid n/2} q^m \sum_{i \mid n/(2m), \text{ odd}} \mu(i).$$ The identity $\sum_{i \mid s}\mu(i)=1_{s=1}$ implies that $\sum_{i \mid n/(2m), \text{ odd}} \mu(i) = 1_{n/{2m} \text{ is a power of 2}}$, and thus, if we let $2^{v_2(n)}$ be the highest power of $2$ dividing $n$, then $$(*) \sum_{d \mid n} a_d d=\sum_{i=0}^{v_2(n)-1} (q^{2^{-(i+1)}})^n.$$
It is not hard to see that the sum $(*)$ cannot be of the form $\sum a_i^n - \sum b_j^n$. (Sketch of Proof: The values of $\alpha_i,\beta_j$ are the poles of $x\frac{A'(x)}{A(x)}$, while $(*)$ implies that $x\frac{A'(x)}{A(x)}$ has infinitely many poles at $q^{2^{-(i+1)}}$.)
[Addendum] Identity $(*)$ also implies that $$A(x) = \prod_{i \ge 1}(1-qx^{2^i})^{-2^{-i}}$$ which leads to the functional equation $$\frac{A(x^2)}{A^2(x)} = 1-qx^2.$$ As for the odd $q$ case, the same arguments lead to $$A(x) = \prod_{i \ge 1}(\frac{1-qx^{2^i}}{1-x^{2^i}})^{-2^{-i}}, \frac{A(x^2)}{A^2(x)} = \frac{1-qx^2}{1-x^2}.$$