Suppose that an injective function $f$ exists. Denote $P(x,y)$ as the assertion that $f(xy)=f(x)+f(y)$. From $P(1,1)$, we have: $$f(1)=2f(1)\implies f(1)=0$$Since $f$ is injective, this means $f(2)\ne0$ and $f(3)\ne0$. From $P(2^n,2)$, we have: $$f(2^{n+1})=f(2^n)+f(2)$$Using this, one can easily show by induction, that $$f(2^n)=nf(2)$$for all integers $n$. From $P(3^n,3)$, we have: $$f(3^{n+1})=f(3^n)+f(3)$$Using this, one can also easily show by induction, that $$f(3^n)=nf(3)$$for all integers $n$. Let $m$ be the smallest positive integer such that $mf(2)$ and $mf(3)$ are integers. Finally, from $P(2^{mf(3)},3^{-mf(2)})$, we have: $$f(2^{mf(3)}\cdot3^{-nf(2)})=f(2^{mf(3)})+f(3^{-mf(2)}$$$$f(2^{mf(3)}\cdot3^{-nf(2)})=mf(3)f(2)-mf(2)f(3)$$$$f(2^{mf(3)}\cdot3^{-nf(2)})=0$$which is a contradiction, because we know that $f(1)=0$, and $2^{mf(3)}\cdot3^{-nf(2)}\ne1$. Therefore, $f$ can never be injective.

The answer is Yes. Let $p_k$ be the $k$th prime number. For every prime $p_k$ and positive rational number $n$, define $\mu_{p_k}(n)=\nu_{p_k}(P)-\nu_{p_k}(Q)$ where $P$ and $Q$ are relatively prime positive integers such that $n=\frac PQ$. One can verify that the following function satisfies the equation: $$f(x)=\mu_{p_1}(x)+\frac{\mu_{p_2}(x)}2+\frac{\mu_{p_3}(x)}3+\frac{\mu_{p_4}(x)}4+\cdots$$Next, we will show that this function is surjective. In other words, we want to show that for any integers $a$ and $b$ such that $b>0$, there always exists $x$ such that $f(x)=\frac ab$. This is true, consider $x=p_b^a$. This would give us $f(p_b^a)=\frac{\mu_{p_b}(p_b^a)}b=\frac ab$.