Well it depends what you mean by a closed form..
$S_n= n! [\frac{1}{1!}+..+\frac{1}{(n-1)!}] \,.$
Then
$en!-S_n= n! [1-\frac{1}{n!}-\frac{1}{(n+1)!}-...]$
$en!-n!+2-S_n=1- n! [-\frac{1}{(n+1)!}-...]$
Now, I think an easy computation shows that the RHS is between 0 and 1.
Thus
$\lfloor en!-n!+2-S_n \rfloor =0$
Hence
$\lfloor en!-n!+2\rfloor =S_n$
Is this a closed form or not? :)
P.S. Got to go, and typed in a hurry thus there might be mistakes in the computation. The basic idea should be right though...