Let $C$ be the category of associative commutative rings with 1 and let $F:C\to C$ be a functor which commutes with the forgetful functor to abelian groups (i.e. $F$ is a functorial way to define another multiplication on every associative commutative ring with 1). Assume also that on $\mathbb Z$ the new multiplication is equal to the old one. Is it true that $F$ is the identity functor?
Yes, $F$ is the identity functor.
The multiplication on $R$ would come in the form of a natural transformation $R \times R \to R$ which is bilinear in each variable. In commutative rings, the functor $R \mapsto R \times R$ is represented by the polynomial algebra $\Bbb Z[x,y]$, and so by the Yoneda lemma natural transformations $R \times R \to R$ are represented by polynomials $f(x,y) \in \Bbb Z[x,y]$. In short, your new multiplication on $F(R)$ must be of the form $x * y = f(x,y)$ for $f$ some polynomial with integer coefficients.
In order for $x * y$ to be distributive, we need $f(x + x',y) \equiv f(x,y) + f(x',y)$, so that $f$ is linear in each variable. This forces $x * y = axy$ for some $a \in \Bbb Z$. However, the only choices of $a$ that ensure $F(R)$ is always unital are $a = \pm 1$, and the only one that preserves the unit of $\Bbb Z$ is $a=1$.