One trivial solution is the constant function $f \equiv 0$. Let $f$ be a nontrivial function for which the conditions (1) and (2) hold. We will show that there exists a natural number $c$ and a prime number $p$ for which it holds that $f(a)=cv_p(a)$ for each $a \in \mathbb{Z}^*$, where $v_p(a):=\text{the exponent of }p\text{ in the canonical factorization of }a$:

let us note at first that $f(1)=f(-1)=0$ (proof:
$$f(1)=f(1 \cdot 1)=f(1)+f(1), \quad f(1)=f((-1) \cdot (-1))=f(-1)+f(-1);$$
from this and from (2) it follows that there exists a prime number $p$ for which $f(p) \ne 0$; for $c:=f(p)$ we will show that $f(a)=cv_p(a)$ holds for every $a \in \mathbb{Z}^*$; namely, for each prime $q \ne p$ there exists nonzero integers $a,\beta$ for which $1=ap+\beta q$, so that the inequality $0=f(ap+\beta q) \ge \min\{f(ap),f(\beta q)\}$ holds; from
$$f(ap)=f(a)+f(p) \ge f(p)=c \ne 0$$
it follows that $f(\beta q)=0$ and $f(q)=0$; let $a=\pm p^kq^\beta r^\gamma...$ be the canonical factorization of $a$; then
$$f(a)=f(\pm p^k)+f(q^\beta)+f(r^\gamma)+\dots=f(\pm 1)+f(p^k)=kf(p)=cv_p(a)$$
It remains to note that each such function satisfies the conditions (1) and (2), and therefore it represents a nontrivial solution to the given problem.