Let's denote the Fibonacci numbers by $F_0=0,F_1=1,F_{n+2}=F_{n+1}+F_n \; \forall n \ge 0$. According to Zeckendorf's theorem, every positive integer can be represented uniquely as the sum of some (at least $1$) distinct Fibonacci numbers where no two consecutive Fibonacci numbers occur.
Suppose positive integers $a=\sum_{r=1}^{p}F_{i_r}$ and $b=\sum_{s=1}^{q}F_{j_s}$ are in their Zeckendorf forms, i.e. $i_1 \ll \cdots \ll i_p,j_1 \ll \cdots \ll j_q$, where $x \ll y \Leftrightarrow x+2 \le y$. Since $F_0=0$ and $F_1=F_2=1$, we can require that $i_1,j_1 \ge 2$ if both $a$ and $b$ are non-zero. Define the Fibonacci product of $a$ and $b$ by $a \circ b=\sum_{r=1}^{p}\sum_{s=1}^{q}F_{i_r+j_s}$ (this result is not necessarily in its Zeckendorf form), and the Zeckendorf form of $0$ to be $0=F_0$. It can be verified that $0$ is the identity element of the Fibonacci product. By a theorem of Knuth, the Fibonacci product on non-negative integers is both commutative and associative (the introduceintroduction of an identity does not break these), which makes the non-negative integers a commutative monoid.
In the same paper, Knuth concluded that Fibonacci product is monotonically increasing in each variable, which means it has the cancellation property. Thus the monoid embeds into its Grothendieck group $G_F$. Consider the submonoid generated by $\{ F_i \}_{i \ne 1}$ we get that $\Bbb{Z} \vartriangleleft G_F$ ($F_i \circ F_j=F_{i+j} \; \forall i,j \ge 2$), but the other elements of $G_F$ are not clear to me (e.g. I don't even know if $G_F$ has any torsion element). Has the group $G_F$ been studied somewhere, or the structure of $G_F$ is too trivial to be looked into? Any referrence or direct description is welcome.
Correction: There's a typo in the "carry rules" $(8)$ & $(9)$ in Knuth's paper. The correct carry rules should be: $$\overline{0(d+1)(e+1)} \rightarrow \overline{1de} \\ \overline{0(d+2)0e} \rightarrow \overline{1d0(e+1)}$$ for $d,e \ge 0$. Knuth seemed to use these correct rules in his following discussion so it did no harm to his conclusion.