Suppose that $R$ is a UFD. Let $F(X) = X^{-n} \sum_{k\ge0} r_k X^k \in K((X))$, with $r_0 \neq 0$. Write $ r_k = \frac{p_k}{q_0\cdot q_1 \cdots q_k}, \quad (k \geq 0),$ with $q_k$ prime with $p_k$ (they are unique up to units).
Suppose that $F \in \mathrm{Frac}(R[[X]])$. Then you can find $A(X) = \sum_{k\ge0} a_n X^k$ and $B(X) = \sum_{k\ge0}b_n X^k$ in $R[[X]]$ such that: $ \sum_{k\ge0}r_k X^k = (\sum_{k\ge0}a_n X^k)(\sum_{k\ge0}b_n X^k)^{-1}.$ You can suppose that $b_0 \neq 0$. This imply that for all $k$: $a_n = \sum_{p+q=k} r_p b_q.$ Multiply this equality by $q_0\cdot q_1 \cdots q_k$, then you deduce that $q_k$ divides $b_0$ (for all $k$).
So a necessary condition for $F(X)$ to be in $\mathrm{Frac}(R[[X]])$ is the following: $(*) \ \ \bigcap_{k \geq 0} q_kR \neq \{0\}, $ i.e. $\gcd(q_0,q_1,\dots) < \infty$. This explains why $\exp(X)$ is not in $\mathrm{Frac}(R[[X]])$ as proved in your link. I expect that $(*)$ is also sufficient, but I am not sure.