For example $S=\{1,2,3,...,n\}$ for $p\not =q, 1\le p,q,\le n$ we have $a=1,b=-p-q, c=pq$ and $a\not =b, a\not =c, b\not =c$.

Rust wrote: For example $S=\{1,2,3,...,n\}$ for $p\not =q, 1\le p,q,\le n$ we have $a=1,b=-p-q, c=pq$ and $a\not =b, a\not =c, b\not =c$. But $b$ does not belong to S! and $c$ doest not always belong to S (for $p=n$ and $q>1$, for example, $c=pq>n$)!

I am sorry, I don't read, that $a,b,c\in S$. In other words, your condition is: let $p\not =q\in S$, then exist $0\not =a\in S$, suth that $-a(p+q)\in S, apq\in S$. If $|p|>1,|q|>1$ it give contradition (we can choose maximal $|p|,|q|, p,q\in S$). Therefore S had no more 4 terms. If S had 4 terms, then $S=-1,0,1,p$. But in these case one of $\pm (p\pm 1)$ don't belong S. Therefore S had no more 3 terms. For example $S=\{-1,0,1\}$ work.