Is it possible to solve the equation $$aF(x)+axF'(x)+bF'(x)+c=0$$ for $x$, where $F(x)$ is a parameter?
It is important to note that this is not a differential equation: $F$ is not an unknown for which we have to solve, rather it is unknown but we consider it fixed, i.e., it is a parameter. For example, if $F(x)=x$ then the solution is $x=-\frac{b+c}{2a}$, but we need a generic solution which works for any $F$ given.
(All we know of of F is that it is a cumulative distribution function.)
If no, is it possible to come up with a meaningful approximation? (Such as something similar to Taylor expansion.)
Observe that $$ aF(x)+axF'(x)+bF'(x)+c = \frac{\mathrm{d}}{\mathrm{d}x} \left( a x F(x) + b F(x) + c x \right) \text{.} $$ Since this is zero, there is a constant, $d$, such that $$ a x F(x) + b F(x) + c x = d \text{.} $$ We now have a choice.
This is as close as one can get to "a generic solution which works for any $F$ given." This is because almost all randomly chosen $F$s do not satisfy the original equation. (For instance, try $F(x) = \dfrac{1}{2} (1- \mathrm{erf}(-x/\sqrt{2}))$, the CDF of the standard normal distribution, which does not satisfy the given differential equation and for which there is no particular $x$ solving the equation unless $a = b = c = 0$.)
(It's been a couple of days and the question has not been corrected. There have been comments made to the Question and this Answer, but none offered concrete improvements or actionable criticism.)
The Question is defective. Describing these defects requires more space than is available in a comment.
First, your example of a CDF is $F(x) = x$. That's no CDF and using this as an example makes it less clear what you mean.
Your equation appears without quantification; consequently, all free variables appearing in it have an implicit universal quantifier. It also contains several undeclared variables, so who knows what they are. This equation cannot possibly hold for all $a$, $b$, and $c$, all $x$, and all $F$, so it is fair for a reader to solve for $F$ to find out what is the compatible domain of the variables.
Finally, the phrase "where $F(x)$ is a parameter" decouples the $x$s appearing in $F(x)$ and $F'(x)$ from the other $x$ appearing in your equation. This makes "$x = \dfrac{-a F(x) - b F'(x) - c}{a F'(x)}$" the literal solution to your question as asked : your question declares that the $x$s on the right-hand side of this solution are not the $x$ on the left-hand side.
So now I'll guess at the intended question.
(Emphasis added to show where $x$ is quantified.)
Another way to go is
Let $$ \hat{U}(x) = \hat{a}F(x)+\hat{a}xF'(x)+\hat{b}F'(x)+\hat{c} \text{.} $$
First, we handle the much simpler case $\hat{a} = 0$. In this case, if there is a zero of $\hat{U}$ it satisfies $F'(x) = -\hat{c}/\hat{b}$, so $x = (F')^{-1}(-\hat{c}/\hat{b})$. Note that this inverse may not exist or may be set-valued for particular choices of $\hat{b}$ and $\hat{c}$. For nonexistence, let $-\hat{c}/\hat{b} < 0$ or $-\hat{c}/\hat{b}$ greater than the maximum of $F'$. For set-valued, consider any unimodal distribution to give $F$ and any $-\hat{c}/\hat{b}$ between the minimum and maximum of $F'$; this gives two zeroes for $\hat{U}$, one on each side of the mode. For even more discrete zeroes, choose a multimodal distribution to produce $F$. Henceforth, we consider only $\hat{a} \neq 0$. (Note: we have shown that $\hat{U}$ need not have a unique zero when $\hat{a} = 0$.)
Since $\hat{a} \neq 0$, we may simplify the given expression. Divide through by $\hat{a}$ to obtain $$ U(x) = F(x) + x F'(x) + b F'(x) + c \text{,} $$ where $b = \hat{b}/\hat{a}$ and $c = \hat{c}/\hat{a}$. Observe that $\hat{U}(x) = 0$ if and only if $U(x) = \hat{U}(x)/\hat{a} = 0/\hat{a} = 0$.
For any $x$ not in the support of $F'(x)$, i.e., any $x$ where $F'(x) = 0$, this is $U(x) = F(x) + c$ and so such an $x$ is a zero of $U$ if $F(x) = -c$. (This is easy to arrange if $0 \leq -c \leq 1$, for abitrarily large sets of $x$. Just let $F$ increase to $-c$, remain constant (forcing $F' = 0$ in this interval) as long as needed, then increase to $1$.) Henceforth, we require $F'(x) > 0$. Since $F(x) = 0$ for all $x$ to the left of the first $x$ where $F'(x) > 0$ (and may continue to do so for larger $x$ if $F'$ is not continuous), we also require $F(x) > 0$. By the same argument applied to the CDF $1-F(-x)$, we require $F(x) < 1$. (Note: we have shown that a zero of $\hat{U}$ need not be unique for $\hat{a} \neq 0$. Consequently, uniquess is hopeless.)
To clarify the nonuniqueness, an example. Let $F$ be the sum of four copies of the CDF of the standardized normal distribution (mean $0$, variance $1$) each scaled by $1/4$, $b = 70$, and $c = -6$. Then a plot of this "4 Normals" $U(x)$ shows $8$ zeroes.
It is easy enough to arrange more. (I suspect it is possible to arrange for the set of zeroes to have a limit point.)
Let $M$ be any upper bound on $F'(x)$. Then, using $0 < F < 1$ and $0 < F' \leq M$, for certain ranges of $c$ we can get (loose) upper or lower bounds on $x$ for which $U(x) = 0$ : \begin{align*} &\text{if $c \leq -1$, } & x &\geq \frac{-c-bM-1}{M} \\ &\text{if $0 \leq c$, } & x &\leq \frac{-c -bM}{M} \end{align*} Notice that from the first of these we may deduce $x \geq -b \ (c \leq -1)$ and from the second $x \leq -b \ (c \geq 0)$, highlighting that $U$ does something interesting around $x = -b$. In fact, $U$ loses dependence on $F'$ at $x = -b$, so if $F(-b)+c = 0$, then $x = -b$ is a zero of $U$. This suggests a small reorganization of $U$ may be useful: $$ U(x) = F(x) + (x+b) F'(x) + c $$ and that we should consider $x < -b$, $x = -b$, and $x > -b$ separately. Since we can trivially verify whether $U(-b) = 0$ once $c$ and $F$ are chosen, we assume $x \neq -b$, henceforth.
Now we can improve the bounds above. \begin{align*} &\text{if $x < -b$ and $c \leq -1$, } & U(x) &< 0 \\ &\text{if $x < -b$ and $-1 < c \leq 0$, } & x &< -b \\ &\text{if $x < -b$ and $0 < c$, } & x &\leq \frac{-c-bM}{M} \\ &\text{if $x > -b$ and $c \leq -1$, } & x &\geq \frac{-c-bM-1}{M} \\ &\text{if $x > -b$ and $-1 < c < 0$, } & x &> -b \\ &\text{if $x > -b$ and $0 \leq c$, } & U(x) &> 0 \text{.} \end{align*}
So perhaps you wonder why we are spending so much time bounding the locations of the zeroes. Because writing a direct formula for the zeroes is hopeless (especially given the vast failure of uniqueness described above). So numerical rootfinding will generally be unavoidable. But what if we choose to use only "simple" CDFs? The standard normal distribution is a "simple" distribution. Using its CDF for $F$, we have $$ U_{\text{std.normal}}(x) = \exp(-x^2/2)(x+b)/\sqrt{2\pi})+(1/2)(1-\mathrm{erf}(-x/\sqrt{2}))+c \text{.} $$ Perhaps for very carefully chosen $b$ and $c$, magical cancellation occurs, but there is no general formula for the zero(-es) in terms of $b$ and $c$ here.