Let
\begin{equation}
M = \begin{pmatrix} a & b \\ c & d\\ \end{pmatrix}
\end{equation}
and consider the following transformation
\begin{equation}
\left[B(M )\right]z = \sum_{ k \in \mathbb{Z}} b_k \left[Mz\right]^k
\end{equation}
on $\mathbb{P}^1\mathbb{(F)}$.
What can we deduce from
\begin{equation}
\left[B(M )\right]z = \left[A(I)\right]z = \sum_{ k \in \mathbb{Z}} a_k \left[Iz\right]^k = \sum_{ k \in \mathbb{Z}} a_k z^k
\end{equation}
where
\begin{equation}
I = \begin{pmatrix} 1 & 0 \\ 0 & 1\\ \end{pmatrix}?
\end{equation}
If we let $\mathbb{F} = \mathbb{C} $ then we can assume that $M$ has one eigenvector; furthermore, in terms of projective geometry, the eigenvector(s) are fixed points since
\begin{equation}
Mz = \lambda z = \lambda \begin{pmatrix} x \\ y \end{pmatrix} \implies M \begin{bmatrix} x : y \end{bmatrix} = \lambda \begin{bmatrix} x : y \end{bmatrix} = \begin{bmatrix} x : y \end{bmatrix} ,
\end{equation} see [Needham ch .3] for a proof. For example
$a=d = 0 \land b=c=1 \implies \begin{bmatrix} 1 : 1 \end{bmatrix},\begin{bmatrix} -1 : 1 \end{bmatrix} $
are fixed points
(i.e. $M \equiv \frac{1}{t}$ has $\pm 1$ as fixed points); this is perhaps a rigorous restatement of your comment
"seemed less about what happened 'at infinity,' and more about what
happened 'away from zero'"
since the defining property of the transformation $M \equiv \frac{1}{t}$ is that it has $\pm 1$ as fixed points as well as the fact that it permutes $\infty$ ($\overset{\text{def}}{=} [1,0]$) with $0$; more generally a Mobius transformation is uniquely defined by its behavior on 3 points (see [Needham ch .3]).
In particular the Mobius transformation $M \equiv \frac{1}{t}$ is "more about what is happening to" the disc $D_1 = \{|z| < 1 \ | \ z \in \mathbb{P}^1\mathbb{(F)} \}$ and the disc $D_2 = \{|z| > 1 \ | \ z \in \mathbb{P}^1\mathbb{(F)} \}$ on the Rieman sphere $\mathbb{P}^1\mathbb{(F)} $ in the sense that the two discs are being permuted; i.e. $M(D_1) = D_2 $ and $M(D_2) = D_1 $.
Since a Mobius transformation preserves conics and in particular preserves discs, a "generic" Mobius transformation (case #\ref{3}) satisfies a similar property by that given in the answer to the post you linked. Let us analyze the generating functions and prove it rigorously.
It's worth considering a degenerate case first
Case #1 $c = 0$
If $c=0 $ then (after relabeling if necessary) $M (z)= (az+b)$ so that $A(z) = B(M(z))$ implies
\begin{equation}
\sum_{ k \in \mathbb{Z}} a_k z^k = \sum_{ k \in \mathbb{Z}} b_k (M (z))^k = \sum_{ k \in \mathbb{Z}} b_k (a z+b)^k = \sum_{ k \in \mathbb{Z}} b_k \sum_{n \in \mathbb{Z}} \binom{k}{n}(az)^nb^{k-n}
\end{equation}
\begin{equation}
\sum_{ k \in \mathbb{Z}} a_k z^k= \sum_{ n \in \mathbb{Z}} b_n \sum_{k \in \mathbb{Z}} \binom{n}{k}(az)^kb^{n-k} = \sum_{ n \in \mathbb{Z}}\left( \sum_{k \leq n } b_n \binom{n}{k}a^kb^{n-k}\right)z^k
\end{equation}
so that
\begin{equation} \label{1} \tag{1} A(z) = B(M(z)) \implies \left( \sum_{k \leq n } b_n \binom{n}{k}a^kb^{n-k}\right)=a_k \end{equation}
The next case
Case #2 $ad-bc = 0$
is trivial since it implies that
\begin{equation} A(z)= B(M(z)) = B\left(\frac{a}{c}\right) \end{equation}
is effectively a constant function in its domain of convergence (and therefore there is an infinite number of "un-interesting" $B$ that satisfy this equation).
Case #3 $$ad-bc\neq 0 \land c \neq 0\label{3} \tag{3}$$ i.e. $M$ is an "interesting" Mobius transformation.
In this case, we have that that $M$ factors through the following functions
- $T_1: z \mapsto z+ \frac{d}{c}$
- $R_1: z \mapsto \frac{1}{z}$
- $R_2: z \mapsto \frac{ad-bc}{c^2}z$
- $T_2: z \mapsto z+ \frac{a}{c}$
(see equation (3) of [Needham ch .3]) so that
\begin{equation}
M= T_2 \circ R_2 \circ R_1 \circ T_1
\end{equation}
so that if we let
\begin{equation}
C(z)= A\left(T_1^{-1}(z)\right)
\end{equation}
and let
\begin{equation}
D(z)= B\left(T_2 \left(R_2 \left(z \right)\right)\right)
\end{equation}
then we have that
\begin{equation} A(z) = B\left(\frac{az+b}{cz+d}\right) \iff \end{equation}
\begin{equation} \iff C(z) =
D\left(\frac{1}{z}\right) \land A(z)= C(T_1 (z)) \land D(z)= B\left(T_2 \left(R_2 \left(z \right)\right)\right) \end{equation}
as was needed to show. In other words, by applying eq.(\ref{1}) we have that it reduces to the following conditions
- $\left( \sum_{k \leq n } c_n \binom{n}{k}\left(\frac{d}{c}\right)^{n-k}\right)=a_k $
- $\left( \sum_{k \leq n } b_n \binom{n}{k}\left(\frac{ad-bc}{c^2}\right)^k\left(\frac{d}{c}\right)^{n-k}\right)=d_k$
- $C(z)$ and $D(z)$ satisfy the property by that given in the answer to the post you linked
One final remark: the property that was given in the answer to the post you linked is just one of many nice properties that rational generating functions satisfy. The property I proved here has a geometric flavor but there are (as the rigorous proof somewhat suggests) many nice combinatorial properties as well. The whole fourth chapter of Stanley is dedicated to rational generating functions. Right off the bat, he gives theorem 4.1.1 which gives four equivalent conditions on rational generating functions; the contender for most straightforward combinatorial interpretation perhaps being condition (ii) which states
\begin{equation}
A(x) = \sum_{n\geq 0} f(n)x^n = \frac{p(x)}{q(x)} \iff (\forall n \in \mathbb{N}) \ f(n+d)+\alpha_1f(n+d-1)+...+\alpha_d f(n) =0
\end{equation}
where $\ \alpha_dx^d+\alpha_{d-1}x^{d-1}+...+\alpha_1 x+1 $ and $\text{deg}(p) \leq \text{deg}(q)$, i.e. that $A(x)$ is the generating function for a linear homogenous recurrence relation.
