$\newcommand\ka\kappa\newcommand{\de}{\delta}\newcommand\ep\varepsilon$The problem obviously reduces to this one: find
\begin{equation}
L:=\lim_{m\to\infty}\frac{S(c,m,n_1,n_2)}{S(c,m-1,n_1-1,n_2-1)},
\end{equation}
where $c:=a/b\ne1$ and
\begin{equation}
S(c,m,n_1,n_2):=\sum_{k=0}^m c^k A_k(m,n_1,n_2),
\end{equation}
where
\begin{equation}
A_k(m,n_1,n_2):=\binom{n_1}k \binom{n_2}{m-k}.
\end{equation}

In accordance with the linked post, assume that $m<n_1$ and $m<n_2$, so that $m\to\infty$,
\begin{equation}
n_1/m\to\nu_1\in[1,\infty),\quad n_2/m\to\nu_2\in[1,\infty).
\end{equation}

By considering the ratios $r_k:=c^{k+1}A_{k+1}(m,n_1,n_2)/(c^k A_k(m,n_1,n_2))$, one sees that
\begin{equation}
S(c,m,n_1,n_2)\sim\sum_{k\sim \ka m} c^k A_k(m,n_1,n_2),
\end{equation}
where
\begin{equation}
\ka:=\frac{c \nu _1-\sqrt{\left(c \nu _1+c+\nu _2-1\right){}^2-4 (c-1) c \nu _1}+c+\nu _2-1}{2 (c-1)}\in(0,1).
\end{equation}
(More specifically, note that $r_k\ge1$ for $k\le k_*$ and $r_k\le1$ for $k\ge k_*$, for a certain integer $k_*\sim\ka m$. Next, note that for each real $\ep>0$ there is some $\de>0$ such that for all large enough $m$ and all $k=0,\dots,m-1$ we have the implication
if $k/m-\ka\ge\de\implies r_k<1-\ep$. So, for integers $k\ge(\ka+3\de)m$ we will have $c^k A_k(m,n_1,n_2)\le(1-\ep)^{\de m}c^{k_*+1} A_{k_*+1}(m,n_1,n_2)$ and hence

$\sum_{k\ge (\ka+3\de)m} c^k A_k(m,n_1,n_2)\le m (1-\ep)^{\de m} c^{k_*+1} A_{k_*+1}(m,n_1,n_2)
=o(\sum_{|k-\ka m|\le3\de m} c^k A_k(m,n_1,n_2))$.

Similarly, $\sum_{k\le (\ka-3\de)m} c^k A_k(m,n_1,n_2)
=o(\sum_{|k-\ka m|\le3\de m} c^k A_k(m,n_1,n_2))$.

Finally here, note that we can choose $\de>0$ to be however small.)

Also, for $k\sim \ka m$,
\begin{equation}
\frac{A_k(m-1,n_1-1,n_2-1)}{A_k(m,n_1,n_2)}
=\frac{(m-k)(n_1-k)}{n_1 n_2}\sim\frac{(1-\ka )(\nu_1-\ka )}{\nu_1\nu_2}.
\end{equation}
So,
\begin{equation}
L=\frac{(1-\ka )(\nu_1-\ka )}{\nu_1\nu_2}.
\end{equation}

Comment 1: $\lim_{c\to1}L_0=\frac{\nu_1+\nu_2-1}{(\nu_1+\nu_2)^2}$, where we have used $\lim_{c\to1}\kappa=\frac{\nu_1}{\nu_1+\nu_2}$ (see Pinelis's answer). This is consistent with the limit for $a=b$ in the above question: $\lim_{c\to1}L_0(c, \nu_1, \nu_2)=L_0(1, \nu_1, \nu_2)$, meaning that $L_0=L_0(c, \nu_1, \nu_2)$ is continuous at $c=1$. $\endgroup$Comment 2: $\lim_{c\to0}L(c, \nu_1, \nu_2)=L(0, \nu_1, \nu_2)=\frac{1}{\nu_2}$, where $\lim_{c\to0}\kappa=0$ (see the definition of $L$ in Pinelis's answer, and we have $L_0=\sqrt{c}L$). Note that $c=0$ means $a=0, b\neq0$. $\endgroup$Comment 3: $\lim_{c\to\infty}L_0(c, \nu_1, \nu_2)=L_0(\infty, \nu_1, \nu_2)=0$: When $c\to\infty$, we have $1-\kappa\sim\sqrt{\nu_2/c}$ for $\nu_1=1$ and $1-\kappa\sim\nu_2/(c(\nu_1-1))$ for $\nu_1>1$, so $(1-\kappa)(\nu_1-\kappa)\sim\nu_2/c$, and finally $L_0\sim1/(\nu_1\sqrt{c})$. Note that $c=\infty$ means $a\neq0, b=0$. $\endgroup$Remark: The summation has the symmetry $S(a,b,m,n_1,n_2)=S(b,a,m,n_2,n_1)$, which leads to the symmetry property of the limit: $L_0(c,\nu_1,\nu_2)=L_0(1/c,\nu_2,\nu_1)$ . And from the above 3 comments, $L_0$ at $c=1,0, \infty$ indeed satisfy this symmetry property. Especially, $L_0\sim\sqrt{c}/\nu_2 (c\to0)$ is equivalent to $L_0\sim1/(\nu_1\sqrt{c}) (c\to\infty)$ because of the symmetry property. $\endgroup$