The formal group law associated with a generating function $f(x) = x + \sum_{n=2}^\infty a_n \frac{x^n}{n!}$ is $$f(f^{-1}(x) + f^{-1}(y)).$$ In my thesis, I found a large number of examples of formal group laws that have combinatorial interpretations and thus have nonnegative coefficients. In Sec 9.1 I conjectured the following characterization for positivity of a formal group law:
Conjecture. $f(f^{-1}(x) + f^{-1}(y))$ has nonnegative coefficients if and only if $$\phi(x) = \frac{1}{\frac{d}{dx} f^{-1}(x)}$$ has nonnegative coefficients.
At least one direction is easy: The positivity of the FGL implies positivity of $\phi(x)$.
I have not been able to prove the converse, but there is some evidence. Start with $$\phi(x) = 1 + t_1x + t_2\frac{x^2}{2!} + t_3\frac{x^3}{3!} + \cdots$$ for indeterminates $t_i$ and define $f(x)$ by $f(0) = 0$, $1/(f^{-1})'(x) = \phi(x)$, or equivalently, $f'(x) = f(\phi(x))$. Then we can compute the coefficients of $f(f^{-1}(x) + f^{-1}(y))$ and they seem to all be polynomials with nonnegative coefficients in the variables $t_i$.
Often it is more illuminating to consider the slightly more general symmetric function $$F = f(f^{-1}(x_1) + f^{-1}(x_2) + \cdots).$$ The expansion of $F$ in the monomial basis of the ring of symmetric functions is
\begin{align*}
F = m_1 &+ (2t_1)\frac{m_{11}}{2!} + (3t_2)\frac{m_{21}}{3!} + (6t_1^2 + 6t_2)\frac{m_{111}}{3!}\\
&+ (4t_3)\frac{m_{31}}{4!} + (12t_1t_2 + 6t_3)\frac{m_{22}}{4!} + (36t_1t_2 + 12t_3)\frac{m_{211}}{4!} \\
&+ (24t_1^3 + 96t_1t_2 + 24t_3)\frac{m_{1111}}{4!} + (5t_4)\frac{m_{41}}{5!} + (30t_2^2 + 30t_1t_3 + 10t_4)\frac{m_{32}}{5!}\\ &+ (60t_2^2 + 80t_1t_3 + 20t_4)\frac{m_{311}}{5!} + (120t_1^2t_2 + 120t_2^2 + 150t_1t_3 + 30t_4)\frac{m_{221}}{5!}\\
&+ (420t_1^2t_2 + 240t_2^2 + 360t_1t_3 + 60t_4)\frac{m_{2111}}{5!} \\
&+ (120t_1^4 + 1320t_1^2t_2 + 480t_2^2 + 840t_1t_3 + 120t_4)\frac{m_{11111}}{5!}\\
&+ \cdots
\end{align*}
Note that $f(x)$ here has a combinatorial interpretation due to Bergeron-Flajolet-Salvy: $f(x)$ is the exponential generating function for increasing trees weighted by their degree sequence in the variables $t_i$. So there is reason to think that there is a combinatorial interpretation of $F$ in terms of increasing trees.
An interesting special case if $\phi(x) = 1 + x^2$, so that $f(x) = \tan(x)$. Then the associated formal group law is a sum of Schur functions of staircase-ribbon shape: $$f(f^{-1}(x_1) + f^{-1}(x_2) + \cdots ) = \sum_{n=1}^\infty s_{\delta_{n} / \delta_{n-2}}$$ where $\delta_n$ is the partition $(n,n-1, n-2, \ldots, 1)$. (See Ardila-Serrano, Prop 3.4.) This can also be interpreted in terms of binary increasing trees.
In many examples given in my thesis I found that there was a combinatorial interpretation of the FGL in terms of chromatic symmetric functions, but I was not able to apply those methods to this more general case.
Edit: Tom Copeland suggested I share some of the Sage code I used to generate these coefficients. Here is a Jupyter notebook in CoCalc that shows the computations.
Best Answer
Given $\phi(x)\in\mathbb{R}[[x]]$, with $\phi(0)=1$, we have defined $g(x):=\int^x_0{dt\over \phi(t)}$, $f:=g^{-1}$ and $$F(x,y)=f\big(g(x)+g(y)\big)=\sum_{n=0}^\infty \psi_n(x) {y^n\over n!}\in\mathbb{R}[[x,y]].$$ Let's write a recursion for the coefficient sequence $\psi_n=\partial_y^nF(x,0)\in\mathbb{R}[[x]]$, solving by series the differential equation satisfied by $F$,
$$\cases{\phi(x)\, F_x(x,y)=\phi(F(x,y))\\ F(x,0)=x\ .}$$
One finds $\psi_0=x$, $\psi_1=\phi,\dots$ . Let's take $\partial_y^n$ at $ {y=0}$ on both sides. Faà di Bruno: $$\partial_y^n\big( \phi\circ F\big)\big|_{y=0}=\Big(\sum_{\alpha\in\operatorname{par}[n]} \phi_y^{(|\alpha|)} (F)\, \prod_{s\in\alpha} \partial_y^{|s|}F \Big) \ \Big|_{y=0} =\sum_{\alpha\in\operatorname{par}[n]} \phi^{(|\alpha|)}(x) \prod_{s\in\alpha} \psi_{|s|}(x) ,$$ (Legenda: The sum is indexed on the set of all partitions of $[n]:=\{1,2,\dots,n\}$, and $|\cdot|$ denotes cardinality. The latter equality comes from $F(x,0)=x$ and $\partial_y^{j}F(x,y)\big|_{y=0}=\psi_j$). Now we isolate the term $\phi'\psi_n$, that corresponds to the partition $\alpha$ into a single class, from the terms of the sum indexed on the set of non-trivial partitions, with $|\alpha|>1$, denoted $\operatorname{par}^*[n]$. Note that each of these terms contains more than one factor $\psi_j$.
$$\phi\psi'_n -\phi' \psi_n =\sum_{\alpha\in\operatorname{par}^*[n]} \phi^{(|\alpha|)} \prod_{s\in\alpha} \psi_{|s|} .$$ Multiplying by the integrating factor $\phi^{-2}$ , and since $\psi_n(0)=0$, for $n>1$ $$ \psi_n(x) =\phi(x)\int_0^x\big(\!\sum_{\alpha\in\operatorname{par}^*[n]} \phi^{(|\alpha|)} \prod_{s\in\alpha} \psi_{|s|}\,\big)\phi^{-2}\ dt .$$
It is now clear by complete induction that for any $n\ge1$, $\psi_n$ is equal to $\phi$ times a series with positive coefficients, proving your conjecture.