Inverting formal power series wrt. composition
You can use Hensel's lemma in the following variant:
Hensel's lemma: Let $A$ be an integral domain which is complete w.r.t. an ideal $I$. Suppose we are given $F \in A[[Y]]$ and $a \in I$ such that $F(a) \equiv 0 \bmod I^s$ for some $s \geq 1$, and $F'(a) \in A^\times$. Then there exists a unique $b \in I$ such that $F(b) = 0$ and $b \equiv a \bmod I^s$.
Given $f(X) = f_1X + \ldots \in R[[X]]$ with $f_1 \in R^\times$ (the dots mean terms of higher order), take $A = R[[X]]$, $F(Y) = f(Y)-X \in R[[X]][[Y]]$, $a = f_1^{-1}X$ and $s = 2$ in the lemma. We have $$F(f_1^{-1}X) = f_1(f_1^{-1}X) + \ldots - X \equiv 0 \mod {X^2},$$ $$F'(f_1^{-1}X) = f_1 + \ldots \in R[[X]]^\times,$$ so Hensel's lemma gives a unique $g \in R[[X]]$ such that $f(g(X)) = X$ and $g(X) = f_1^{-1}X +\ldots$. The same argument with $g$ instead of $f$ gives a unique $h(X) = f_1X + \ldots \in R[[X]]$ such that $g(h(X)) = X$. Now $$f(X) = f(g(h(X))) = h(X),$$ so $h=f$ and thus $f(g(X)) = g(f(X)) = X$.
Edit 11/2018: I think you can prove the lemma as follows. Replacing $F(Y)$ with $F(Y+a)F'(a)^{-1}$ we can assume $a = 0$ and $F'(0) = 1$. Then we put $x_0 = 0$ and $x_{n+1} := x_n - F(x_n)$. Then $x_n \equiv 0 \bmod I^s$ for all $n \geq 0$ by induction. We claim: $$x_{n+1} \equiv x_n \bmod I^{n+s} \quad \text{for all $n \geq 0$}.$$ The case $n=0$ is clear. Assume $x_n \equiv x_{n-1} \bmod I^{n+s-1}$ for the induction. We can write $$F(Y) - F(Z) = (Y-Z)(1+YG(Y,Z)+ZH(Y,Z))$$ for certain $G, H \in A[[Y,Z]]$. We find $$F(x_n) - F(x_{n-1}) \equiv x_n - x_{n-1} \bmod I^{n+s},$$ hence $x_n - F(x_n) \equiv x_{n-1} - F(x_{n-1}) \bmod I^{n+s}$, or also $x_{n+1} \equiv x_n \bmod I^{n+s}$. The claim shows that $(x_n)$ is a Cauchy sequence. By completeness, it converges to some $b \in I^s$. It satisfies $b = b-F(b)$, hence $F(b) = 0$. For uniqueness, assume $F(b) = F(c) = 0$ with $b,c \in I^s$. We find $$0 = (b-c)(1+bG(b,c)+cH(b,c)),$$ which implies $b = c$ since $1+I \subseteq A^\times$. It seems that the assumption of $A$ to be integral is not necessary.
I'm going to basically copy a proof given by Cartan.
It is sufficient for there to exist a unique $g$ that $f(0) = 0$ and $f'(0) \ne 0$, as you have said. We also know that $f_1g_1$ must equal $1$.
We want $f\circ g(X) = X$, So we try to find the conditions for which the coefficient of $X^n$ in $f\circ g$ is $0$. We know that this is equal to the coefficient of $X^n$ in $$f_1g+f_2g^2+\cdots + f_ng^n\tag{1}$$ So we have a condition $$f_1g_n + P_n(f_2,\dots,f_n,g_1,\dots,g_{n-1})= 0\tag{2}$$ Where $P_n$ is a polynomial that may be calculated from $(1)$. Because $f_1 \ne 0$, $f_1g_1 = 1$ determines $g_1$. We then take $(2)$ with $n = 2$ to find $g_2$, and by induction we can find all $g_n$.
We therefore have the existence and uniqueness of such an inverse compositional series.
We can also show that $f$ is the only series for which $g$ is an inverse. Construct $f'$ as we constructed $g$ such that $f'(0) = 0$ and $g\circ f' = X$. We have that $$f' = X\circ f' =f\circ g \circ f' = f\circ X = f$$
This doesn't explicitly construct the coefficients, but proves that they may be constructed, which is a little more elegant.