Factorial over Superfactorial!

Note first that we can represent $S$ as

$S = \displaystyle\sum_{n=1}^{\infty} \dfrac{n!}{\prod_{k=0}^{n} k!} = \sum_{n=1}^{\infty} \dfrac{1}{\prod_{k=0}^{n-1} k!} = \dfrac{1}{0!} + \dfrac{1}{1!} + \dfrac{1}{1! \times 2!} + \dfrac{1}{1! \times 2! \times 3!} + ....,$

since $0! = 1.$ Thus $S$ and $S_{0}$ are identical except for the term $\dfrac{1}{0!} = 1,$ and so $S - S_{0} = 1 = 1*e^{0}.$ Thus $A + B = 1 + 0 = \boxed{1}.$

For completeness, (at the suggestion of the Challenge Master), note that

$\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{\prod_{k=1}^{n} k!} \lt \sum_{n=1}^{\infty} \dfrac{1}{k!} = e - 1,$

and that the series of partial sums $\displaystyle\sum_{n=1}^{N} \dfrac{1}{\prod_{k=1}^{n} k!}$ is monotonically increasing. Thus we can conclude that the series $S_{0}$ converges, and since $S = 1 + S_{0}$ we also know that $S$ converges.

Won't this do?

$\begin{aligned} S & = \dfrac{1}{1!} + \dfrac{1\times 2}{1!\times2!} + \dfrac{1\times 2\times 3}{1!\times2!\times3!} + \dfrac{1\times 2\times 3\times 4}{1!\times2!\times3!\times4!} + ... \\ & = \dfrac{1}{1!} + \dfrac{2!}{1!\times2!} + \dfrac{3!}{1!\times2!\times3!} + \dfrac{4!}{1!\times2!\times3!\times4!} + ... \\ & = \dfrac{1}{1!} + \dfrac{1}{1!} + \dfrac{1}{1!\times2!} + \dfrac{1}{1!\times2!\times3!} + ... \\ & = 1 + S_0 \end{aligned}$

$\Rightarrow S-S_0 = 1 = 1\times e^0\quad \Rightarrow A+B = 1+0 = \boxed{1}$