Notation:
Let $$Q_1[f(x)] = \lim _{h \rightarrow 1} \frac{f(x + h) - f(x) }{h}$$
And let
$$Q_1^{-1}\left[f(x)\right] = G(x)|Q_1\left[g(x)\right] = f(x)$$
Consider the equation
$$a_0(x) + a_1(x)f(x) + a_2(x)Q_1\left[f(x)\right] = 0$$
How do we express the general solution to this equation?
A natural first step is to divide the equation by $a_2(x)$ to find:
$$\frac{a_0(x)}{a_2(x)} + \frac{a_1(x)}{a_2(x)}f(x) + Q_1\left[f(x)\right] = 0$$
We can rename the coefficients (being that they are general) to:
$$ b_0(x) + b_1(x) f(x) + Q_1\left[f(x)\right] = 0 $$ Now subtract $b_0(x)$ from both sides: (I am attempting a Duhamel's formula style approach)
$$b_1(x)f(x) + Q_1\left[f(x)\right] = -b_0(x) $$
Now note that for a pair of functions there exists a product rule. Given $a(x)$ and $b(x)$
$$Q_1\left[a(x)b(x)\right] = Q_1\left[a(x)\right]Q_1\left[b(x)\right] + Q_1\left[a(x)\right]b(x) + a(x)Q_1\left[b(x)\right]$$
If we let $b(x) = f(x)$ then we note that we obtain a form of:
$$\left(Q_1\left[a(x)\right] + a(x)\right)Q_1\left[f(x)\right] + Q_1\left[a(x)\right]f(x)$$
At this point it is evident that to solve:
$$b_1(x)f(x) + Q_1\left[f(x)\right] = -b_0(x) $$
We need a function $\lambda(x)$ such that
$$\lambda(x)b_1(x)f(x) + \lambda(x)Q_1\left[f(x)\right] = Q_1\left[f(x)a(x)\right] $$
Therefore:
$$\lambda(x)b_1(x) = Q_1\left[a(x)\right]$$
$$\lambda(x) = Q_1\left[a(x)\right] + a(x)$$
Thus:
$$\lambda(x)b_1(x) + a(x) = \lambda(x)$$
Therefore:
$$\lambda(x) = \frac{a(x)}{1 - b_1(x)}$$
And thus both equations breakdown to:
$$ \frac{b_1(x)}{1 - b_1(x)} a(x) = Q_1\left[a(x)\right] $$
Let $a(x) = 2^{g(x)}$ then it follows
$$Q_1\left[2^{g(x)}\right] = 2^{g(x)}\left(2^{Q_1\left[g(x)\right]} - 1\right) $$
Thus we are merely tasked with solving
$$\left(2^{Q_1\left[g(x)\right]} - 1\right) = \frac{b_1(x)}{1 - b_1(x)} $$
Which has solution
$$g(x) = Q_1^{-1}\left[\log_2\left\lbrace\frac{1}{1 - b_1(x)}\right\rbrace\right] $$
Thus the general solution has the form
$$2^\left({Q_1^{-1}\left[\log_2\left(\frac{1}{1 - b_1(x)}\right)\right]}\right)f(x) = Q_1^{-1}\left[-b_0(x)*\frac{2^{Q_1^{-1}\left[\log_2\left(\frac{1}{1 - b_1(x)}\right)\right]}}{1 - b_1(x)}\right]$$
Therefore
$$f = \frac{Q_1^{-1}\left[-b_0(x)*\frac{2^{Q_1^{-1}\left[\log_2\left(\frac{1}{1 - b_1(x)}\right)\right]}}{1 - b_1(x)}\right]}{2^{Q_1^{-1}\left[\log_2\left(\frac{1}{1 - b_1(x)}\right)\right]}}$$
Simplifying and substituting $\frac{a_0(x)}{a_2(x)}$ for $b_0(x)$ and $\frac{a_1(x)}{a_2(x)}$ for $b_1(x)$
We conclude that the general solution to:
$$a_0(x) + a_1(x)f(x) + a_2(x)Q_1\left[f(x)\right] = 0$$
is:
$$f = \frac{Q_1^{-1}\left[\frac{a_0(x)}{a_2(x)}*\frac{2^{Q_1^{-1}\left[\log_2\left(\frac{1}{1 - \frac{a_1(x)}{a_2(x)}}\right)\right]}}{\frac{a_1(x)}{a_2(x)} - 1}\right]}{2^{Q_1^{-1}\left[\log_2\left(\frac{1}{1 - \frac{a_1(x)}{a_2(x)}}\right)\right]}} $$
I think this is correct. But a second eye is always helpful