Necessary and sufficient condition to be completely multiplicative
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If $f(1)\ne 0,1$ then $f\ast f(1)= f(1)^2\ne f(1)=f(1)\tau(1)$.
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If $f(1)= 0$ then take the least $n$ such that $f(n)\ne 0$, you'll have $f\ast f(n)=0\ne f(n)\tau(n)$.
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So assume that $f(1)=1$.
Take the least $n$ such that $f(n)\ne \prod_{p^k\| n} f(p)^k$.
For all $m$ let $g(m)=\prod_{p^k\| m} f(p)^k$.
then $$f\ast f(n)-f(n)\tau(n) = g\ast g(n) + f(n)-g(n)-f(n)\tau(n)$$ $$= g(n)\tau(n)+f(n)-g(n)-f(n)\tau(n)= (g(n)-f(n))(\tau(n)-1)\ne 0$$