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 $p$.
UPDATED. 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 $$S_p:=\sum_{i_p} f(n_p-1, i_{p-1}, i_p) = [\tfrac{x_p^{n_p-1}}{(n_p-1)!}\tfrac{y^{i_{p-1}+1}}{(i_{p-1}+1)!}]\ (1+e^{x_p}(e^{-y}-1))^{-1}$$ and correspondingly \begin{split} S_{p-1} &:= \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}}}{i_{p-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)!}]\ \left.\tfrac{\partial}{\partial t} (1+e^{x_p}(e^{-t}-1))^{-1} \right|_{t=-\log(1+e^x(e^{-y}-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)!}]\ (e^{x_p}-1)g_{p-1}^{-2} + g_{p-1}^{-1}, \end{split} where $$g_k := 1+e^{x_k+\dots+x_p}(e^{-y}-1).$$ In general, $S_k$ (ie. the sum over $i_k,\dots,i_p$) is given by a linear combination of $g_k^{k-p-1}, g_k^{k-p}, \dots, g_k^{-1}$ with coefficients depending on $x_{k+1},\dots,x_p$ but not on $y$. Furthermore, $S_k$ can be obtained from $S_{k+1}$ by replacing each $g_{k+1}^{-\ell}$ with $\ell(e^{x_{k+1}+\dots+x_p}-1)g_k^{-\ell-1} + \ell g_k^{-\ell}$.
Then the iterated sum in question is obtained as the coefficient of $\tfrac{x_1^{n_1-1}}{(n_1-1)!}\cdots \tfrac{x_p^{n_p-1}}{(n_p-1)!}\tfrac{y^{m+1}}{(m+1)!}$ in $S_1$. It may be possible that $S_1$ has a simpler description but I have not found one yet.