$\nexists y \in l^1$ such that $\forall x \in S: L(x) = \sum\limits_{n\ge 1}(x y)\lbrack n \rbrack$
Take $x=(0,0,...,0,1,0,0...)$ which has $1$ in the $k-$th place and $0$ elsewhere. We get $0=Lx=\sum x[n]y[n]=y[k]$. Thus $y[k]=0$ and this is true for all $k$. But then $Lx=0$ for all $x$ which is contradiction since $L(1,1,1...) \neq 0$.