Grothendieck group of the Fibonacci monoid 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 introduction 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.
 A: This operation is not mysterious at all! The monoid $(\mathbf N,\circ)$ is isomorphic to a multiplicative submonoid $T$ of the commutative ring $\mathbf Z[\varphi] = \mathbf Z[t]/(t^2-t-1)$, where $\varphi = \tfrac{1+\sqrt{5}}{2}$ is the golden ratio. In particular, it is isomorphic to a submonoid of $\mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$, where $\mathcal P$ is the set of prime ideals in $\mathbf Z[\varphi]$, and the first factor is the powers of the fundamental unit $\varphi$. In fact, the groupification $T^{\text{gp}} \hookrightarrow \mathbf Q(\varphi)^\times \cap \mathbf R_{>0} \cong \mathbf Z \oplus \mathbf Z^{\oplus \mathcal P}$ is an isomorphism; see the corollary below.
I first wrote this as an answer with full details, but a small literature search (starting from Knuth's paper on MathSciNet and looking for forward references) turned up some original sources that are a little slicker than what I wrote. So in the interest of readability, I have removed some of the proofs.
The main reference is the 2-page paper [Arnoux, 1989], explaining the relation to $\mathbf Z[\varphi]$. In addition, [Zhuravlev, 2007] notes that every nonzero element of $\mathbf Z[\varphi]$ can be written (non-uniquely) as $\pm\varphi^{-n}t$ for $t \in T$ and $n \in \mathbf N$, giving the promised computation of $T^{\text{gp}}$ of the corollary below.
Notation. Let $f \colon \mathbf Z[t] \twoheadrightarrow \mathbf Z[\varphi]$ be the quotient map $t \mapsto \varphi$, and let $g \colon \mathbf Z[\varphi] \to \mathbf Z$ be the group homomorphism $a+b\varphi \mapsto b$. Note that $g$ is not a ring homomorphism; in fact $g(\varphi^n) = F_n$ by Binet's formula. Write $h$ for the composition $g \circ f$; this exhibits a polynomial $\sum a_it^i$ as a radix-F expansion of its image $\sum a_i F_i$.
The Fibonacci product $\circ$ has an obvious extension to $\mathbf N \times \mathbf N \to \mathbf N$ by setting $n \circ 0 = n = 0 \circ n$ for all $n \in \mathbf N$ (which does not affect associativity). We view $\mathbf Z[\varphi]$ as a subring of $\mathbf R$ in the obvious way, and it has a conjugation $\overline{(-)} \colon \mathbf Z[\varphi] \to \mathbf Z[\varphi]$ taking $\varphi$ to $1-\varphi = \varphi^{-1} = \tfrac{1-\sqrt{5}}{2}$.
Definition. Define the subset $Z \subseteq t^2\mathbf Z[t]$ of elements of the form $P(t)=\sum_{i=2}^r a_it^i$ with all $a_i \in \{0,1\}$ and $a_ia_{i+1} = 0$ for all $i$. We remove $0$ from this set and add $1$, since this will be our multiplicative unit. Let $T \subseteq \mathbf Z[\varphi]$ be the image of $Z$ under $f$. Then Zeckendorf's theorem shows that $f$ and $g$ give bijections
$$Z \stackrel\sim\to T \stackrel\sim\to \mathbf N.$$
By definition, the map $h$ has the property $h(xy) = h(x)h(y)$ for all $x,y \in Z$. But $Z$ is not closed under multiplication, so this doesn't produce a monoid isomorphism $h \colon Z \stackrel\sim\to \mathbf N$. The key point is:
Proposition [Arnoux, 1989] The set $T$ is closed under multiplication. In particular, $g \colon T \to \mathbf N$ is a monoid isomorphism for the Fibonacci product on $\mathbf N$, i.e.
$$g(xy) = g(x) \circ g(y)\qquad \text{for all } x, y \in T.$$
This also gives a conceptual proof of associativity of $(\mathbf N,\circ)$ (which is not used in the proof). So we see that $\mathbf Z[\varphi]$ interpolates nicely between $\mathbf Z[t]$ and $(\mathbf N,\circ)$.
The proposition can easily be shown by hand using the multiplicative structure of $f$ together with Lemma 3 of Knuth; see the revision history of this post for such a proof. Instead, Arnoux deduces it from the following:
Lemma [Arnoux, 1989]. The set $T \setminus \{1\}$ is given by the elements $t=a+n\varphi$ with $a,n \in \mathbf Z_{>0}$ such that $\overline{\!\ t\ \!} \in (\varphi-2,\varphi-1)$. The map $g^{-1}$ is given by $n \mapsto a_n + n\varphi$, where
$$a_n = \left\lfloor (n+1)\tfrac{-1+\sqrt{5}}{2} \right\rfloor$$
is Hofstadter's $G$-sequence (OEIS A005206).
See Lemmas 2 and 3 in Arnoux. The final statement is not there, but can easily be deduced from the first. This shows that $T$ is closed under multiplication as $(\varphi-2,\varphi-1) \subseteq (-1,1)$. In addition, it gives the clean formula
$$n \circ m = nm + na_m + ma_n.$$
Note that in the first statement, we don't need the assumption $a > 0$. Indeed, if $a \leq 0$ we get $\overline{\!\ t\ \!} = a+n\overline\varphi \leq \overline\varphi < \varphi-2$ since $\overline\varphi < 0$ and $n \geq 1$.
Finally, we need the following observation that appears to be due to [Zhuravlev, 2007]; see Proposition 4.1.
Lemma. Every element $x \in \mathbf Z[\varphi]$ can be written as $\pm\varphi^{-n} \cdot t$ for some $t \in T$ and $n \in \mathbf N$.
Since $T$ and $\varphi$ are positive, the sign agrees with the sign of $x$. The proof (and the whole paper) is notationally heavy (and logically hard to follow), so let me include an argument here.
Proof. Since $\lvert \overline\varphi \rvert < 1$, we get $\pm\overline{\!\ t\ \!} \in (\varphi-2,\varphi-1)$ for $n \gg 0$. If $t = a+n\varphi$, we necessarily have $n \neq 0$, for otherwise $t$ and therefore $x$ is $0$. Wihout loss of generality, we may assume $n > 0$. By the previous lemma (and the discussion after), we get $t \in T$, so $x = \pm\varphi^{-n} \cdot t$ as desired. $\square$
(Notational note: what Zhuravlev calls $\delta(n)$ is related to my $g^{-1}(n)$ via $-\delta(n)/\varphi = \overline{g^{-1}(n)}$. Zhuravlev's Fibonacci sequence is off by $1$ compared to Knuth.)
Corollary. There exists a choice of generators of each prime ideal of $\mathbf Z[\varphi]$ inducing an injection $\psi \colon T \hookrightarrow \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$. For any such choice, the map $\psi^{\text{gp}} \colon T^{\text{gp}} \to \mathbf Z \oplus \mathbf Z^{\oplus \mathcal P}$ is an isomorphism, and the quotient map $T \to \mathbf N^{\oplus \mathcal P}$ is surjective.
Proof. We saw that $T$ is a submonoid of $(\mathbf Z[\varphi]\setminus\{0\},\times)$. Since $\mathbf Z[\varphi]$ is a real quadratic principal ideal domain, we can produce an isomorphism $(\mathbf Z[\varphi]\setminus\{0\},\times) \cong \mathbf Z/2 \oplus \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$ by picking generators for each prime ideal. By the lemma, we may pick a generator in $T$ for each prime ideal. Since all elements of $T$ and all chosen representatives are positive, we don't need the sign factor $\mathbf Z/2$, giving an embedding $T \hookrightarrow \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$. The final two statements follow immediately from the lemma since $\varphi^n \in T$ for all $n \geq 2$. $\square$
While subgroups of free abelian groups are free, the same does not hold for commutative monoids, i.e. there is no unique factorisation into irreducible elements. For instance, the elements $\varphi^n$ for $n \geq 2$ and $n=0$ are in $T$, but $\varphi$ is not (as $g(\varphi) = 1 = g(\varphi^2)$ and $g$ is injective on $T$). So $\varphi^6$ factors both as $(\varphi^2)^3$ or $(\varphi^3)^2$. The result above is probably the most precise you are going to get (also because it depends on infinitely many choices).
Note that negative powers of $\varphi$ are not in the image. In fact, $(\mathbf N,\circ)$ is a sharp monoid: if $a,b \neq 0$, then $a \circ b \neq 0$. I don't know if there exists a choice of generators for which no element picks up a negative power of $\varphi$ (i.e. the image is in $\mathbf N \oplus \mathbf N^{\oplus \mathcal P} \subseteq \mathbf Z \oplus \mathbf N^{\oplus \mathcal P}$), but this seems unlikely to me.

References.
[Arnoux, 1989]  P. Arnoux, Some remarks about Fibonacci multiplication. Appl. Math. Lett. 2.4, p. 319-320 (1989). DOI:10.1016/0893-9659(89)90078-5
[Zuravlev, 2007]  V. G. Zhuravlev, Sums of squares over the Fibonacci $\circ$-ring. Zap. Nauchn. Semin. POMI 337, p. 165-190 (2006). Translation in J. Math. Sci., New York 143.3, p. 3108-3123 (2007). DOI:10.1007/s10958-007-0195-1
