$F$ 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: $$F(C)=F(A_{0})+F(B_{0})+F(A_{1})+F(B_{1})$$
 A: 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 $\,F(x) = 0\,$.
A: If you apply the functional equation to $1\cdot x + 1\cdot y$, you get
$$F(x+y) = F(x) + F(y) + 2F(1),$$
so we're awfully close to $F$ being linear. Let's use $\alpha$ to denote $F(1)$, and speculate that we can find $k$ such that $G(x) = F(x) + k\alpha$ is linear. Plugging $G$ into the above equation, we get
$$G(x+y) - k\alpha = G(x) - k\alpha + G(y) - k\alpha + 2\alpha$$
By assumption, the $G$'s drop out, and we get $k=2$.
So define $G$ as $$G(x) = F(x) + 2\alpha,$$ and then $G$ satisfies $$G(x+y) = G(x) +G(y).$$
If you require $F$ to be continuous, then so is $G$, and this famous functional equation has the only solutions $G(x) = G(1) \cdot x$. This follows even if we only allow positive $x$, which is a restriction I advocated for in the comments.
So, putting things back in terms of $F$,
$$F(x) = 3\alpha x - 2\alpha$$
This gives us $F(1) = \alpha, F(2) = 4\alpha$, and $F(5) = 13\alpha$. Now if we apply the functional equation to $5 = 2\cdot 2 + 1\cdot 1$, we get
$$13\alpha = 4\alpha + 4\alpha + \alpha + \alpha,$$
forcing $\alpha$ to be $0$, and thus $F$ is identically equal to $0$.
(Note that in their answer dxiv accomplishes the same thing without requiring continuity, so their result is stronger. But I like the method of finding a perturbed version of $F$ that is linear, so I'll keep this answer up.)
