Here, $e = \sum_{k=0}^\infty 1/k!$. Define $S_n$ as:
\begin{align} S_n = n!e - n!\sum_{k=0}^n \frac{1}{k!} \end{align} where $n!\sum_{k=0}^n \frac{1}{k!}$ is an integer. Write $e = 1/0!+1/1!+...+1/n!+1/(n+1)!+...$ \begin{align} S_n &= n!\left(e - \sum_{k=0}^n \frac{1}{k!}\right) = n!\left( \frac{1}{0!}+ \frac{1}{1!}+...+\frac{1}{n!}+\frac{1}{(n+1)!} +... - \sum_{k=0}^n \frac{1}{k!}\right)\\ &=n!\left( \frac{1}{0!}+ \frac{1}{1!}+...+\frac{1}{n!}+\frac{1}{(n+1)!} +... - \frac{1}{0!} - \frac{1}{1!} - ... - \frac{1}{n!} \right)\\ &=n!\left( \frac{1}{(n+1)!}+\frac{1}{(n+2)!}+... \right) \\ &= \frac{1}{(n+1)}+\frac{1}{(n+2)(n+1)}+... < \frac{1}{(n+1)}+\frac{1}{(n+1)^2}+... = \frac{1}{n} \end{align}
So, $0<S_n<1/n$. Now, if $e=p/q$ then we would have an integer between $0$ and $1$ for a large $n$.
$$0 < n!p - qn!\sum_{k=0}^n \frac{1}{k!}<\frac{q}{n}$$
Is this correct?
It's basically fine. To tie up the details, you should specify exactly how large $n$ needs to be.
Also, writing "$0<S_n =$ (something that isn't equal to $S_n$) is bad notation. I'm not even sure exactly what you were trying to say there. If you're not sure how to write what you mean in symbols, use some words to clear things up.