The following variant of the proof may be simpler: the relation $F(y)=[x^n] \frac{1}{1+x}\frac{1}{(1-\frac{x}{1+x}(1-y))^{n+1}}$ shows that $F(y)=(-1)^n F(1-y)$. Thus $F(1)-F(y)=(-1)^n \sum_{k=1}^n { n\choose k} { n+k \choose k} (-1)^{k+1} (1-y)^{k}$ and (using the Beta-integral $\int_0^1 (1-y)^{k-1} y^n\,dy=\frac{(k-1)!\,n!}{(n+k)!}$) $$\int_0^1 \frac{1 -F(y)}{1-y}\,y^n\,dy=(-1)^n \sum_{k=1}^n{n \choose k}\frac{(-1)^{k+1}}{k}=(-1)^n\,H_n$$

Please note the obvious typo, should be ${Z_n \over m^n}$ instead of ${Z_n \over m}$ above.

$F_n$ ``counts'' the no. $Z_n$ of individuals in the $n-$th generation of a Galton-Watson process with reproduction function $F(s)=qs+ps^2$. Thus (look up the literature) $\mathbb{E} Z_n=m^n$, $\mathrm{Var}(Z_n)=\sigma^2 m^{n-1}{m^n-1 \over m-1}$ where $m=\mathbb{E}(Z_1)=1+p, \sigma^2={\mathrm{Var}}(Z_1)=pq$. Further $${Z_n \over m}\longrightarrow W\;\;\mbox{a.s.}$$ where $W$ is positive on $\{Z_n\longrightarrow \infty\}$, and the Laplace transform $\ell$ of $W$ is characterized by the the equation $\ell(mt)=q\ell(t)+p\ell(t)^2$, and right derivative $-\ell^\prime(0)=1$ in $0$.

(1) I have corrected some inaccuracies in the first part (apologies for being confusing), expanded it and hope it is sufficiently clear now. (2) In your post above the index $b$ should be $n$.

But note that in the asymmetric case $$\mathbb{P}(Y_n\geq r)=\mathbb{P}(X_n\geq r)+({p \over q})^r \mathbb{P}(X_n\leq -(r+1))$$ for $r\geq 0$. Thus for the cdf you essentially only need the cdf of the binomial distribution (incomplete Beta function).

Apologies, my statement for $p\not={1/2}$ above is false. (The derivation can be done as in Feller, but the sum doesn't simplify). It seems that only in the symmetric case the distribution of $Y_n$ has a simple expression.

It's Thm 1 in III.7 of Feller I. For $p\not={1 \over 2}$ replace $2^{-n}$ with $p^kq^{n-k}$ , $k:=\lceil{n+r+1 \over 2}\rceil$.

Put $z=x\cdot A^3(x)$, i.e. $x={z \over A^3(x)}$. By Lagrange Inversion $b_0=a_0=1$ and $b_n=[z^n] A(x)=[t^n] {-1 \over 3n-1}{1\over A^{3n-1}(t)}$ for $n\geq 1$. Now use Thm VIII.8 in ``Analytic Combinatorics''.

I haven't found your paper. Is it on the arXiv yet?