As it was noted in another answer, we have $$f(n,m,i) =[\tfrac{x^n}{n!}\tfrac{y^{m+1}}{(m+1)!}]\ \frac{\big(-\log(1+e^x(e^{-y}-1))\big)^i}{i!}.$$
The linked answer essentially establishes the same recurrence for $f(n,m,i)$ as satisfied by $a(n)$, and the conjecture then easily follows by induction on $p$.
Now, let's simplify the given iterated sum expression.
Since $\log(1+e^x(e^{-y}-1))$ is a multiple of $y$, it follows that all summation bounds can be replaced with 0 and $+\infty$, respectively. Then $$\sum_{i_p} f(n_p-1, i_{p-1}, i_p) = [\tfrac{x_p^{n_p-1}}{(n_p-1)!}\tfrac{y^{i_{p-1}+1}}{(i_{p-1}+1)!}] (1+e^{x_p}(e^{-y}-1))^{-1}$$ and correspondingly \begin{split} \sum_{i_{p-1}} & f(n_{p-1}-1, i_{p-2}, i_{p-1}) \sum_{i_p} f(n_p-1, i_{p-1}, i_p) \\ &= [\tfrac{x_{p-1}^{n_{p-1}-1}}{(n_{p-1}-1)!} \tfrac{x_p^{n_p-1}}{(n_p-1)!}\tfrac{y^{i_{p-2}+1}}{(i_{p-2}+1)!}\tfrac{t^{i_{p-1}+1}}{(i_{p-1}+1)!}]\ (1+e^{x_p}(e^{-t}-1))^{-1} \tfrac{\big(-\log(1+e^x(e^{-y}-1))\big)^{i_{p-1}+1}}{(i_{p-1}+1)!} \\ &= [\tfrac{x_{p-1}^{n_{p-1}-1}}{(n_{p-1}-1)!} \tfrac{x_p^{n_p-1}}{(n_p-1)!}\tfrac{y^{i_{p-2}+1}}{(i_{p-2}+1)!}]\ (1+e^{x_p}((1+e^{x_{p-1}}(e^{-y}-1))-1))^{-1} \\ &= [\tfrac{x_{p-1}^{n_{p-1}-1}}{(n_{p-1}-1)!} \tfrac{x_p^{n_p-1}}{(n_p-1)!}\tfrac{y^{i_{p-2}+1}}{(i_{p-2}+1)!}]\ (1+e^{x_{p-1}+x_p}(e^{-y}-1))^{-1}. \end{split}
Continuing similarly, we obtain $$\sum_{i_1,\dots, i_p} \ldots = [\tfrac{x_1^{n_1-1}}{(n_1-1)!}\cdots \tfrac{x_p^{n_p-1}}{(n_p-1)!}\tfrac{y^{m+1}}{(m+1)!}]\ (1+e^{x_1+\cdots+x_p}(e^{-y}-1))^{-1}.$$
The derived generating function further implies the following explicit formula: $$a(n) = \sum_{k=0}^{m+1} k^{n_1+\dots+n_p-p} \binom{n_1+\dots+n_p-p}{n_1-1,\dots,n_p-1}^{-1} k! (-1)^{m+1} \left\{m+1\atop k\right\}.$$