= Solution
\b[No.] Use the <prime-sign automorphism of the multiplicative integers>
$$
s(0)=0,\qquad s(n)=(-1)^{v_2(n)}n\quad(n\ne0),
$$
where $v_2(n)$ is the exponent of two in $|n|$, a <P-adic valuation>. For nonzero integers, $v_2(mn)=v_2(m)+v_2(n)$, so $s(mn)=s(m)s(n)$; the identity also holds when a factor is zero. The exponent of two is unchanged by changing the sign, hence $s(s(n))=n$. Thus $s$ is a multiplicative <structure automorphism>.
It fixes one and sends two to minus two. The addition-graph point $(1,1,2)$ is sent to $(1,1,-2)$, outside the graph of addition. By <automorphism invariance of parameter-free definability>, \b[addition is not definable in the multiplicative reduct].
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