 Solution

No. Use the prime-sign automorphism of the multiplicative integerswhere $v_2(n)$ 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, addition is not definable in the multiplicative reduct.