Some general results that might be interesting.
As the OP said, generating functions do seem like a good method for this kind of recurrences.
Let's consider a more general homogeneous second order linear recurrence with coefficients linear in $n$:
$$\left(A_2 n+B_2 \right) a_{n+2}+\left(A_1 n+B_1 \right) a_{n+1}+\left(A_0 n+B_0 \right) a_n =0$$
Now let's introduce some function defined by its Frobenius series:
$$f(x)=\sum_{n=0}^\infty b_n x^{n+s}$$
Which (we claim) obeys a second order ODE in the form:
$$(\alpha_2 x^2+\beta_2 x+\gamma_2) f''(x)+(\beta_1 x+\gamma_1) f'(x)+\gamma_0 f(x)=0$$
We have:
$$f'(x)=\sum_{n=-1}^\infty (n+1+s) b_{n+1} x^{n+s}$$
$$x f'(x)=\sum_{n=0}^\infty (n+s) b_n x^{n+s}$$
$$f''(x)=\sum_{n=-2}^\infty (n+1+s)(n+2+s) b_{n+2} x^{n+s}$$
$$x f''(x)=\sum_{n=-1}^\infty (n+s)(n+1+s) b_{n+1} x^{n+s}$$
$$x^2 f''(x)=\sum_{n=0}^\infty (n-1+s)(n+s) b_n x^{n+s}$$
Substituting the series, we obtain the following recurrence relation:
$$\alpha_2 (n-1+s)(n+s) b_n+\beta_2 (n+s)(n+1+s) b_{n+1} + $$ $$ + \gamma_2 (n+1+s)(n+2+s) b_{n+2} +\beta_1 (n+s) b_n+\gamma_1 (n+1+s) b_{n+1}+ $$ $$ +\gamma_0 b_n=0$$
Or:
$$\gamma_2 (n+s+1)(n+s+2) b_{n+2}+ $$ $$ +\left( \beta_2 n^2+(\beta_2 (2s+1)+\gamma_1)n+(\beta_2 s+\gamma_1) (s+1) \right) b_{n+1}+ $$ $$ +\left( \alpha_2 n^2+(\beta_1+(2s-1) \alpha_2)n+\alpha_2 s (s-1)+\beta_1 s+\gamma_0 \right) b_n=0 $$
To get rid of quadratic terms, we need to be able to factor out one of the brackets before the first term, let's pick $(n+s+1)$. Which means that the other two coefficients should have the form:
$$q(n+s+1)(n+p)=q(n^2+(p+s+1) n+p(s+1))$$
For the second term we have:
$$q=\beta_2 \\ p=s+\frac{\gamma_1}{\beta_2}$$
For the third term:
$$q=\alpha_2 \\ p=s+\frac{\beta_1}{\alpha_2}-2=\frac{\alpha_2 s (s-1)+\beta_1 s+\gamma_0}{\alpha_2 (s+1)}$$
We have obtained an equation which allows us to get rid of one parameter, for example, $\gamma_0$.
Now our recurrence relation is reduced to:
$$\gamma_2 (n+s+2) b_{n+2}+ \left(\beta_2n+\beta_2s+\gamma_1 \right) b_{n+1} +\left(\alpha_2 n+\alpha_2(s-2)+\beta_1 \right) b_n=0$$
We now have a linear second order recurrence with six parameters left, which (at least in principle) can be related to the original parameters $A_2,B_2,A_1,\dots$, and we can even set $a_n=b_n$.
In the example from Integral representation for the solution of a difference equation we have:
$$\gamma_2=1 \\
s=0 \\
\beta_2=\gamma_1=-2 \\
\alpha_2=1 \\
\beta_1=3 \\ \gamma_0=1$$
Which makes the generating function obey the following ODE:
$$(x-1)^2 f''(x)+(3 x-2) f'(x)+f(x)=0 \\ f(0)=1, \qquad f'(0)=0$$
Wolfram Alpha gives the exact solution:
$$f(x)=\frac{1}{1-x} \exp \left(- \frac{x}{1-x} \right)$$
From the explicit recurrence form given in the linked question, we also have:
$$f(x)= e \int_0^\infty e^{-(1-x) t} J_0 (2 \sqrt{t} ) dt$$
But this is correct and confirmed numerically.
So, the method allows us to obtain an ODE for a certain class of second order linear recurrence relations with variable coefficients linear in $n$.
Note: for the case in the OP the method doesn't seem to work as stated, some modifications need to be made.
Update: Another example.
Using hypergeometric equation, we can derive the following result. The sequence with explicit form:
$$a_n = \frac{(\alpha)_n}{(\beta)_n} {_2 F_1} (n+1, n+\alpha; n+\beta; p )$$
obeys the recurrence relation:
$$p(p-1) (n+2) a_{n+2}+ [(2p-1) n +(\alpha+2) p -\beta] a_{n+1} + (n+ \alpha) a_n=0 $$ $$ a_0 = {_2 F_1} (1, \alpha; \beta; p ) $$ $$ a_1 = \frac{\alpha}{ \beta} {_2 F_1} (2, 1+\alpha; 1+\beta; p )$$
$p \neq 0, p \neq 1, \qquad \beta \notin \mathbb{Z}$.
For general initial conditions the solution also exists, but will be more complicated.