Here is a slightly different approach which might be helpful. We start with a few notational conventions which help to simplify calculations.
We consider $a,b,c$ and $d$ as functions with domain $\mathbb{N}$ with e.g. $a(n):= a_n$ for $n\in\mathbb{N}$. It is also convienient to consider an expression $(a+b\Delta)$ as operator acting on a function $y$ via
\begin{align*}
(a+b\Delta)y(n)&=a(n)y(n)+b(n)\left(y(n+1)-y(n)\right)\\
&=a_ny_n+b_n(y_{n+1}-y_n)
\end{align*}
We use the shift operator $\operatorname{E}$, defined as $\operatorname{E}y(n):=y(n+1)$ which is connected with the $\Delta$-operator via
\begin{align*}
\Delta&=\operatorname{E}-1\\
\Delta y(n)&=y(n+1)-y(n)\\
&=y_{n+1}-y_{n}\\
(\operatorname{E}-1) y(n)&=\operatorname{E}y(n)-y(n)\\
&=y(n+1)-y(n)
\end{align*}
Now we are ready to go.
We have to find conditions among which the operator
\begin{align*}
a+b\Delta\qquad\text{and}\qquad c+d\Delta
\end{align*}
commute. We obtain
\begin{align*}
(a+b\Delta)(c+d\Delta)&=(c+d\Delta)(a+b\Delta)\\
(a+b(\operatorname{E}-1))(c+d(\operatorname{E}-1))&=(c+d(\operatorname{E}-1))(a+b(\operatorname{E}-1))\tag{1}\\
(a-b+b\operatorname{E})(c-d+d\operatorname{E})&=(c-d+d\operatorname{E})(a-b+b\operatorname{E})\tag{2}\\
(a-b)d\operatorname{E}+b\operatorname{E}(c-d)+b\operatorname{E}d\operatorname{E}
&=d\operatorname{E}(a-b)+(c-d)b\operatorname{E}+d\operatorname{E}b\operatorname{E}\tag{3}\\
\end{align*}
Comment:
Now we apply both sides of (3) to the function $y$. We obtain
\begin{align*}
&\left((a-b)d\operatorname{E}\right)y(n)+\left(b\operatorname{E}(c-d)\right)y(n)
+\left(b\operatorname{E}d\operatorname{E}\right)y(n)\\
&\qquad=\left(d\operatorname{E}(a-b)\right)y(n)+\left((c-d)b\operatorname{E}\right)y(n)
+\left(d\operatorname{E}b\operatorname{E}\right)y(n)\\
&(a_n-b_n)d_ny_{n+1}+b_n(c_{n+1}-d_{n+1})y_{n+1}+b_nd_{n+1}y_{n+2}\\
&\qquad=d_n(a_{n+1}-b_{n+1})y_{n+1}+(c_n-d_n)b_ny_{n+1}+d_nb_{n+1}y_{n+2}\tag{4}
\end{align*}
We observe from (4) the operators commute if and only if the coefficient functions of $y_{n+1}$ and $y_{n+2}$ are equal on both sides. Comparing the coefficients of $y_{n+2}$ we see
\begin{align*}
b_nd_{n+1}=d_nb_{n+1}\\
\end{align*}
from which condition $\frac{d_n}{b_n}=\frac{d_{n+1}}{b_{n+1}}=\text{const.}$ follows.
Using this relationship when comparing the coefficients of $y_{n+1}$ in (4) we obtain
\begin{align*}
(a_n-b_n)d_n+b_n(c_{n+1}-d_{n+1})&=d_n(a_{n+1}-b_{n+1})+(c_n-d_n)b_n\\
a_nd_n+b_nc_{n+1}&=d_na_{n+1}+c_nb_n\\
b_n(c_{n+1}-c_n)&=d_n(a_{n+1}-a_n)\\
b_n\Delta c_n&=d_n\Delta a_n
\end{align*}
and the claim
\begin{align*}
\frac{d_n}{b_n}=\frac{\Delta c_n}{\Delta a_n}=\text{const.}
\end{align*}
follows.
