Does this define a commutative group? I've had a look at a binary law between analytic germs of functions, but I don't know if it is a group law, or where to find any reference.
Let $\mathcal{H}$ be the space of germs of holomorphic functions $f$ at $0$ such that $f(z) \sim z$ at $0$. Alternatively, one can see $\mathcal{H}$ as the space of sequences $(a_n) \in \mathbb{C}^\mathbb{N}$ such that $a_0 = 0$, $a_1 = 1$, and $\limsup_{n \to + \infty} |a_n|^\frac{1}{n} < + \infty$. I'll denote by $\rho$ the radius of convergence of any such germ. For simplicity, I'll endow $\mathcal{H}$ with the topology of uniform convergence on suitable compact neighborhoods of $0$ (that's not optimal, but it will spare me a few headaches for now).
For $f(z) = \sum_{n \geq 0} a_n z^n$ and $g(z) = \sum_{n \geq 0} b_n z^n$, let:
$$f \diamond g (z) := \sum_{n \geq 0} a_n g(z^n).$$
The $n$th coefficient of $f \diamond g$ is given by $\sum_{d | n} a_d b_{n/d}$. In particular, this operation is commutative. let $\rho^* := \min \{1, \rho\}$. Then this formula also gives bounds such as:
$$\rho (f \diamond g) \geq \min \left\{ \sqrt{\rho^* (f)} \rho^* (g), \sqrt{\rho^* (g)} \rho^* (f) \right\} > 0.$$
Since $f \diamond g (z) \sim z$, this binary law preserves $\mathcal{H}$. a small computation show that the $n$th coefficient of $(f \diamond g) \diamond h$
is given by:
$$\sum_{\substack{d_1,d_2,d_3 \leq n \\ d_1 d_2 d_3 = n}} a_{d_1} b_{d_2} c_{d_3},$$
so the law is associative. A neutral element is given by $z$.
There are a few nice things you can do with this. For instance, let $\mathcal{P}$ be the set of primes. For any integer $m \geq 2$, let $\varphi_m (z) := \sum_{n \geq 0} z^{m^n}$. Then the fundamental theorem of arithmetics can be reformulated as:
$$\diamond_{p \in \mathcal{P}} \varphi_p (z)= \frac{z}{1-z}.$$
By the way, this look a lot like the usual identity involving Riemann's Zeta function. Anyway, my next question is: does this law has inverses? A short computation shows that the inverse of $(z-z^m)$ is $\varphi_m$. I think that we can generalise this computation to show that any polynomial has an inverse, by using a linear recurrence on $\mathbb{N}^k$ for a suitable $k$ (depending on the polynomial). By taking limits (or equivalently with some kind of linear recurrence on $\mathbb{N}^\mathcal{P}$), I also think that we can find an inverse for any element of $\mathcal{H}$.
However, I don't see how to prove or disprove that any such inverse has a positive radius of convergence. So, my question could be reformulated as: does any element in $\mathcal{H}$ has an inverse with positive radius of convergence?
If that were false, I could restrict myself to formal series (without the condition that they have a positive radius of convergence), but that wouldn't be fun, and I would lose the formula $f \diamond g (z) := \sum_{n \geq 0} a_n g(z^n)$.
Of course, if it is possible to express the law $\diamond$ from a well-known group law (most likely addition or multiplication), then that would be an acceptable solution.
 A: 
Anyway, my next question is: does this law has inverses?

Yes, it does. Without caring about convergence for the moment, the coefficients $(b_n)$ of the inverse of $f(z) = \sum_{n=1}^\infty a_n z^n$ are given by $b_1 = 1$ and
$$b_n = -\sum_{\substack{d\mid n \\ d < n}} b_d\cdot a_{n/d}$$
for $n > 1$, which shows that we have a uniquely determined sequence of coefficients, and then it only remains to see that the formal power series
$$g(z) = \sum_{n=1}^\infty b_n z^n$$
has positive radius of convergence.
Since $f$ is holomorphic in a neighbourhood of $0$, there is a $K\in [1,\infty)$ such that
$$\lvert a_n\rvert < K^n$$
for all $n\in \mathbb{N}\setminus \{0\}$. We next show that
$$\lvert b_n\rvert < (2K)^n.\tag{1}$$
That is clear for $n = 1$, and for $n > 1$, we estimate
\begin{align}
\lvert b_n\rvert &\leqslant \sum_{\substack{d\mid n\\ d < n}} \lvert a_{n/d}\rvert\cdot \lvert b_{d}\rvert\\
&\leqslant K^n + \sum_{\substack{d\mid n\\1 < d < n}} K^{n/d} (2K)^d \tag{Ind. hyp.}\\
&\leqslant K^n + K^n \sum_{\substack{d\mid n\\1 < d < n}}2^d \tag{$\ast$}\\
&\leqslant K^n\sum_{d=1}^{n-1}2^d\\
&= (2^n-2)K^n\\
&< (2K)^n,
\end{align}
where in $(\ast)$ we used that
$$\frac{n}{d} + d \leqslant n$$
for all divisors $2 \leqslant d \leqslant \frac{n}{2}$ of $n$.
From $(1)$, it follows that $g$ is holomorphic at least in the disk around $0$ with radius $\frac{1}{2K}$, hence $\diamond$ makes $\mathcal{H}$ a group.
