Suppose that we have the Taylor expansions : $$f(x)=\sum_{n=1}^{\infty}\frac{a_{n}}{n!}x^{n}$$ $$g(x)=\sum_{n=1}^{\infty}\frac{b_{n}}{n!}x^{n}$$ Then we have the standard result : $$g(f(x))=\sum_{n=1}^{\infty}\left(\sum_{k=1}^{n}b_{k}B_{n,k}(a_{1},..,a_{n-k+1})\right)\frac{x^{n}}{n!}$$ Where $B_{n,k}(\cdot)$ are the partial Bell polynomials.
Is there any similar result adding one constant term to each one, so with $F(x)=f_0+f(x)$ and $G(x)=g_0+g(x)$?
In fact G(x) does not presents any real complexity added, my question is mainly centered on F(x).
If you expand $$\sum_{n=1}^{\infty} \frac{b_n}{n!} \left( a_0 + a_1 x + \cdots \right)^n$$ formally and collect terms with the same of power of $x$, you'll see that there are infinitely many terms for each power of $x$. Thus you need to deal with issues of convergence. For example, the constant term becomes $$ \sum_{n = 1}^{\infty} \frac{b_n a_0^n}{n!}. $$
Therefore, in the setting of formal power series, $G(F(x))$ is only well-defined if $F(0) = 0$ (i.e. $F$ has no non-zero constant term). Adding a non-zero constant term to $G$ is, as you observe, not a problem at all.