$F$ such that $F(A_{0}B_{0}+A_{1}B_{1})=F(A_{0})+F(B_{0})+F(A_{1})+F(B_{1})$? We know that the only way of turning a multiplication into addition is by a logarithmic function; so if $$C=A_{0}B_{0}$$ then we know that: $$\log(C)=\log(A_{0})+\log(B_{0})$$ but can we extend this idea? . 

In the case of:
$$C=A_{0}B_{0}+A_{1}B_{1}$$
can we prove or disprove the existing of a function F(x) such as: $$F(C)=F(A_{0})+F(B_{0})+F(A_{1})+F(B_{1})$$

No such function exists, except the identically zero function. With the notation $\,f_k = F(k)\,$:
$$
\begin{align}
2 = 1 \cdot 1 + 1 \cdot 1 \quad\implies\quad f_2 &= 4 f_1
\\ 3 = 2 \cdot 1 + 1 \cdot 1 \quad\implies\quad f_3 &= 3 f_1 + f_2 = 7 f_1
\\ 4 = 2 \cdot 1 + 2 \cdot 1 \quad\implies\quad f_4 &= 2f_1 + 2 f_2 = 10 f_1
\\ 7 = 3 \cdot 2 + 1 \cdot 1 \quad\implies\quad f_7 &= 2 f_1 + f_2 + f_3\color{red}{ = 13 f_1}
\\ 7 = 4 \cdot 1 + 3 \cdot 1 \quad\implies\quad f_7 &= 2 f_1 + f_3 + f_4\color{red}{ = 19 f_1}
\end{align}
$$
Since $\,F(7)\,$ can only take one value, it follows that $\,13f_1=19f_1\,$ $\,\implies f_1 = F(1) = 0\,$, then $\,F(k)=0\,$ for $\,k \in \mathbb N\,$ by induction, and therefore for all $\,k \ge 2\,$:
$$
\require{cancel}
k\cdot x = (k-1) \cdot x + 1 \cdot x \;\;\implies\;\; F(k \cdot x) = \cancel{F(k-1)} + F(x) + \cancel{F(1)} + F(x) = 2 F(x)
$$
But then:
$$
\begin{align}
F(2x) &= F( 2 \cdot x) = 2 F(x)
\\ F(4x) &= F(4 \cdot x) = 2 F(x)
\\ F(4x) &= F(2 \cdot 2x) = 2 F(2x) = 4 F(x)
\end{align}
$$
The last two equalities imply $\,2 F(x) = 4 F(x)\,$, so in the end $\,F(x) = 0\,$.
