I start my question with an example that I did. Give the following series $R(x):=\sum_{d=1}^{\infty}(\sum_{l=0}^{d-1} (-1)^{l}\frac{1}{ \ \ell!(d-1-\ell)!})q_{1}^{d} x^d$. I could decompose this as a product of two series $P(x):=\sum_k (-1)^k\frac{1}{k!}q_1^{k} x^k$ and $Q(x):= \frac{1}{k!}q_1^{k} x^k$.
Now I want to see if such decomposition exists for the following series which is the general case of the series $R(x)$. $$Z(x):= \sum_{d=1}^{\infty}(\sum_{l=1}^{d}(-1)^l\sum_{m_1 +2m_2=d}\frac{1}{m_1 2^{m_{2}}}\sum_{i_2=0}^{m_2}\frac{1}{i_2!(m_2 -i_2)!}\frac{1}{(\ell-1-2(m_2 - i_2))!(d-\ell-2i_2)!}q_1^{m_1} q_2 ^{m_2} )x^d .$$ Though $q_1$ is redundant for $R(x)$ I include it to show it's the special case of $Z(x)$ when we put $q_2=0$. What would be the corresponding $P(x)$ and $Q(x)$ ?