Compute $S_n=\sum\limits_{a_1,a_2,\cdots,a_n=1}^\infty \frac{a_1a_2\cdots a_n}{(a_1+a_2+\cdots+a_n)!}$
I did not check your computation thoroughly, but your idea of resummation followed by the use of Stirling numbers definitely seems to work.
Alternatively, here is another answer in finite sum:
$$ \sum_{a_1, \cdots, a_n \geq 1} \frac{a_1 \cdots a_n}{(a_1 + \cdots + a_n)!} = \left( \sum_{k=0}^{n}\binom{n}{k} (-1)^{n+k-1} \sum_{j=0}^{n+k-1} \frac{(-1)^j}{j!} \right) e. \tag{1} $$
This formula also tells that the sum is always a rational multiple of $e$.
Proof of (1). We begin by recalling the multivariate beta identity. Let
$$\Delta^{n-1} = \{ (x_1, \cdots, x_n) \in \mathbb{R}^n : x_i \geq 0 \text{ and } x_1 + \cdots + x_n = 1 \}$$
denote the $(n-1)$-simplex. Then
$$ \mathrm{B}(a_1,\cdots,a_n) := \int_{\Delta^{n-1}} \left( \prod_{k=1}^{n} x_k^{a_k-1} \right) \, dx_1 \cdots dx_{n-1} = \frac{\Gamma(a_1)\cdots\Gamma(a_n)}{\Gamma(a_1+\cdots+a_n)}. $$
This is essentially equivalent to the usual beta identity. Now denoting the sum by $S_n$, we have
\begin{align*} S_n &= \sum_{a_1,\cdots,a_n \geq 1} \frac{1}{a_1+\cdots+a_n} \left( \prod_{k=1}^{n} \frac{a_k}{\Gamma(a_k)} \right) \mathrm{B}(a_1, \cdots, a_n) \\ &= \sum_{a_1,\cdots,a_n \geq 1} \left( \int_{0}^{1} u^{a_1+\cdots+a_n-1} \, du \right) \left( \prod_{k=1}^{n} \frac{a_k}{\Gamma(a_k)} \right) \left( \int_{\Delta^{n-1}} \left( \prod_{k=1}^{n} x_k^{a_k-1} \right) \, d\mathrm{x}\right) \\ &= \int_{\Delta^{n-1}} \int_{0}^{1} \left( \prod_{k=1}^{n} \sum_{a_k=1}^{\infty} \frac{(u x_k)^{a_k-1} a_k}{(a_k - 1)!} \right) \, u^{n-1} du d\mathrm{x}. \end{align*}
The inner sum can be easily computed by the formula
$$ \sum_{a=0}^{\infty} \frac{a+1}{a!}z^a = e^z (1+z), $$
and hence we obtain
\begin{align*} S_n &= \int_{\Delta^{n-1}} \int_{0}^{1} u^{n-1} e^u (1 + ux_1)\cdots(1 + ux_n) \, du d\mathrm{x} \\ &= \sum_{I \subset \{1,\cdots,n\}} \int_{0}^{1} u^{n+|I|-1} e^u \left( \int_{\Delta^{n-1}} \prod_{i\in I}x_i \, d\mathrm{x} \right) \, du \\ &= \sum_{k=0}^{n} \binom{n}{k} B(\underbrace{2,\cdots,2}_{k\text{ terms}}, \underbrace{1,\cdots,1}_{n-k\text{ terms}}) \int_{0}^{1} u^{n+k-1} e^u \, du. \end{align*}
Evaluating the last sum gives the expression $\text{(1)}$ as desired.
Addendum. I checked the preface of the book and found the quote:
Each chapter contains a section of difficult problems, motivated by other problems in the book, which are collected in a special section entitled “Open problems”....
That being said, they are truly intended as exercises for readers!