A basic fact about rings, is that under addition, they form an abelian group.
In this case, the abelian group has order 3, so either of $q$ or $r$ is a generator.
Since the map $a \mapsto -a$ is a group isomorphism for any abelian group (in particular it maps a generator to a generator), it must be the case that either:
$-q = r$, or $-q = q$.
If $-q = q$, then $q + q = p$, in which case $q$ has order 2. But 2 does not divide 3, so this violates Lagrange.
Therefore, $-q = r$, that is $r + q = p$.
So from our given equation:
$r\ast(r + q) = r\ast r + r\ast q$ we get:
$r \ast p = r\ast r + r \ast q$ (from the above)
$r \ast p = r\ast r + r$ (since $q$ is the multiplicative identity).
Now, to see that $r \ast p = p$, note that:
$r \ast p = r\ast(p + p)$ (since $p + p = p$, since it is the additive group identity)
$= r \ast p + r\ast p$. Hence:
$p = -(r \ast p) + r\ast p = -(r\ast p) + r\ast p + r\ast p = p + r\ast p = r\ast p$.
Thus (continuing our first argument):
$p = r\ast r + r$, so that $r\ast r$ is the additive inverse of $r = -q$.
So $r\ast r = -(-q) = q$ (in a group, the inverse of an inverse is the original element).
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The above is tedious, and complicated by notation. In point of fact, there is no reason not to denote $p$ by $0$, and $q$ by $1$. Thus:
$S = \{0,1,r\}$.
It is plain to see we must have $r = 1 + 1$. Now we can just calculate:
$(1+1)(1+1) = (1+1)+(1+1) = [(1+1)+1] + 1 = 0 + 1 = 1$
because $(1+1)+1 = 3\cdot 1 = 0$ (the "3rd power" of 1, written additively) by Lagrange.