Consider the following function where $A=\mathbb{Q}$
$$F(x)=\begin{cases} 2^x & x=A_1\\ x^2 & x=A_2\\ \text{Undefined} & \text{Everywhere Else} \end{cases} $$
$$A_1=\left\{\frac{2m+1}{2n+1}:m,n\in\mathbb{Z}\right\}$$ $$A_2=2\mathbb{Z}$$ $$A_1\subset\mathbb{Q}, \ A_2\subset\mathbb{Q}$$
Say we find the integral $F(x)$
$A_1$ is dense and $A_2$ is countably infinite. (For the time being $A_2$ is countably infinite but later I will take an example where $A_2$ is dense). The Lebesgue measure of both sets is zero but the rest of the function is undefined.
However, $A_1$ is so dense in $\mathbb{R}$ we ignore all points except the discontinuities at $\mathbb{2Z}$. It seems without $A_2$, the integral exists for all real numbers. Hence we need a measure where one set has a measure of $1$ and another set has a measure of $0$.
This is where Folner sequences come into play. $A_1$ and $A_2$ are subsets of rational numbers. There are several Folner Sequences of rational numbers (shown below).
$$m,n\in\mathbb{Z}, \ k\in\mathbb{N}, \ j\in\text{odd}, \ a,b\in\mathbb{R}$$
$$F_{1,r}=\left\{\frac{m}{n}:a<\left|\frac{m}{n}\right|<b,|n|<r,\gcd(m, n)=1\right\}$$
$$F_{2,r}=\left\{\frac{m}{2^k j}:a<\left|\frac{m}{2^k j}\right|<b,|k|<\log_2(r),|j|<r,\gcd(m,2^k n)=1\right\}$$
$$F_{2,r,l}=\left\{\frac{m}{2^k j}:a<\left|\frac{m}{2^k j}\right|<b,|k|<l,|j|<r,\gcd(m,2^k n)=1\right\}$$
To be converted into densities with respect to Folner Sequences of $\mathbb{Q}$ which for subset $X$, we denote as $D_{\mathbb{Q}}(X)$ (specifically $D_{\mathbb{Q}\cap[a,b]}(X)$ where $X$ is $A_1$ or $A_2$); the Densities must follow two properties. For $T_1,T_2\subset\mathbb{Q}$
$$T_1=T_2 \implies D_{\mathbb{Q}\cap[a,b]}(T_1)=D_{\mathbb{Q}\cap[a,b]}(T_2)$$
$$T_1\subseteq T_2 \implies D_{\mathbb{Q}\cap{[a,b]}}(T_1)\le D_{\mathbb{Q}\cap{[a,b]}}(T_2)$$
The densities are
$$D_{\mathbb{Q}\cap[a,b]}(X)=\lim_{r\to\infty}\frac{|X\cap F_{1,r}|}{|F_{1,r}|}$$ $$D_{\mathbb{Q}\cap[a,b]}(X)=\lim_{r\to\infty}\frac{|X\cap F_{2,r}|}{|F_{2,r}|}$$ $$D_{\mathbb{Q}\cap[a,b]}(X)=\lim_{(r,l)\to\infty}\frac{|X\cap F_{2,r,l}|}{|F_{2,r,l}|}$$
Now we can use a technique to compare $D_{\mathbb{Q}\cap[a,b]}(A_1)$ to $D_{\mathbb{Q}\cap[a,b]}(A_2)$ as an "informal measure". We will denote this as $D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(X)$.
$$D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(X)=\lim_{r\to\infty}\frac{|X\cap F_{1,r}|}{|A_1\cap F_{1,r}|+|A_2\cap F_{1,r}|}$$ $$D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(X)=\lim_{r\to\infty}\frac{|X\cap F_{2,r}|}{|A_1\cap F_{2,r}|+|A_2\cap F_{2,r}|}$$ $$D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(X)=\lim_{(r,s)\to\infty}\frac{|X\cap F_{2,r,l}|}{|A_1+F_{2,r,l}|+|A_2+F_{2,r,l}|}$$
We can simplify the densities into
$$F'_{1}=\left\{\frac{m}{n}:a<\left|\frac{m}{n}\right|<b,\gcd(m, n)=1\right\}$$
$$F'_{2}=\left\{\frac{m}{2^k j}:a<\left|\frac{m}{2^k j}\right|<b,\gcd(m,2^k n)=1\right\}$$
$$F'_{2}=\left\{\frac{m}{2^k j}:a<\left|\frac{m}{2^k j}\right|<b,\gcd(m,2^k n)=1\right\}$$
$$D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(X)=\lim_{r\to\infty}\frac{\sum\limits_{0<n<r}|X\cap F'_{1}|}{\sum\limits_{0<n<r}|A_1\cap F'_{1}|+\sum\limits_{0<n<r}|A_2\cap F'_{1}|}$$ $$D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(X)=\lim_{r\to\infty}\frac{\sum\limits_{0<k<\log_2(r)}\sum\limits_{0<j<r}|X\cap F'_{2}|}{\sum\limits_{0<k<\log_2(r)}\sum\limits_{0<j<r}|A_1\cap F'_{2}|+\sum\limits_{0<k<\log_2(r)}\sum\limits_{0<j<r}|A_2\cap F'_{2}|}$$ $$D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(X)=\lim_{(r,l)\to\infty}\frac{\sum\limits_{0<k<s}\sum\limits_{0<j<r}|X\cap F'_{2}|}{\sum\limits_{0<k<s}\sum\limits_{0<j<r}|A_1\cap F'_{2}|+\sum\limits_{0<k<s}\sum\limits_{0<j<r}|A_2\cap F'_{2}|}$$
It seems for any Density with respect to the folner sequences of $\mathbb{Q}$, $D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(A_2)=0$, making $D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(A_1)=1$, hence
$$\int_{D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(A_1\bigcup A_2)}F(x)=\int_{D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(A_1)}F(x)+\int_{D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(A_2)}F(x)=\int 2^x=\ln(2)(2^{b}-2^{a})$$
Am I correct?
Using this approach it seems only few pairs of subsets have a measure and an integral. My belief is that if one set has a measure of $1$ than an integral exists.
If sets have a measure between 0 and 1, for all densities, then it can be used for an average on the function where the sets are defined. So far the only subsets I think are possible is $A_1=\left\{\frac{2m+1}{2n+1}\right\}$ and $A_2=\left\{\frac{2m}{2n+1}\right\}$.
Perhaps we can extend these notions to sets such as $\mathbb{Q}^{\mathbb{Q}}$, $\left\{\frac{\ln(m)}{\ln(n)}\right\}$, $\sqrt{m}-\sqrt{n}$ and so on.
Now what happens when
$$A_1=\left\{\frac{2m+1}{2n+1}:m,n\in\mathbb{Z}\right\}$$
$$A_2=\left\{\frac{2p+1}{2q}:p,q\in\mathbb{Z}\right\}$$
Edit: I have a feeling the Density I defined can only be defined for finite and countably infinite non-dense sets to a dense set. Two countably infinite Dense sets cannot be compared by their denominator but rather their numerator if the sets of the denominators they're defined on (for reduced elements) are equal.
Another Method
Perhaps we can reduce the number of Folner Sequences used for $\mathbb{Q}$ if we take the lowest common denominator of first $\lfloor r \rfloor$ denominators of the rational numbers. We get the following
$$p=\left\{\text{Set of Primes}\right\}$$ $$S(r)=\prod_{p \le r}p^{\lfloor\log_p(r)\rfloor}$$
Giving the sequence
$$F_{3,r}=\left\{a<\frac{m}{S(r)}<b:m\in\mathbb{Z}\right\}$$
However, the Folner Sequence makes us consider elements beyond the first $\lfloor r \rfloor$ denominators for every $r$. If we only considered elements with the first $\lfloor r \rfloor$ denominators (set $V$), we get
$$V\cap F_{3,r}=F_{1,r}$$
If we leave $V$ the same, it closely resembles
$$F_{2,r}$$
I have trouble choose which Folner Sequence works best $F_{1,r}$ or $F_{2,r}$?
If $F_{2,r}$ is the only possible Folner Sequence then if
$$A_1=\left\{\frac{2m+1}{2n+1}:m,n\in\mathbb{Z}\right\}$$
$$A_2=\left\{\frac{2p+1}{2q}:p,q\in\mathbb{Z}\right\}$$
$D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(A_2)=1$ and $D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(A_1)=0$ making
$$\int_{D_{\left\{A_1\bigcup A_2 \right\}\cap[a,b]}(A_1 \bigcup A_2)}F(x)=\int_{D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(A_1)}F(x)+\int_{D_{\left\{A_1\bigcup A_2\right\}\cap[a,b]}(A_2)}F(x)=\frac{1}{3}b^3-\frac{1}{3}a^3$$
I will try and give a high level answer.
Given a Folner sequence $(F_n)$ on a countable group $G$ and a subset $A\subset G$, we can try and define how big A is "according to the Folner sequence" as follows: $$ D(A)= \lim_{ n\rightarrow \infty} \frac{|F_n \cap A|}{ |F_n|}$$. Unfortunately, this object is somewhat badly behaved. For one, this density does not give us a measure in the conventional sense, because countable additivity is not satisfied. For another, it is unclear whether the limit even exists. It is possible to extend this density function so that it is well-defined for any $A\subset G$ using some fancy tools from analysis (Hahn-Banach theorem, ultrafilters, etc), and in this case the extended version of $D$ becomes a translation-invariant, finitely additive probability measure. It is possible to develop a theory of integration for such objects, in line with Lebesgue integration, but due to the failure of countable additivity, basically all of the useful theorems about Lebesgue integrals are false in this context. Moreover, the fancy tools required to get our finitely additive probability measure are all non-constructive, which is to say that we can assert that the integral $\int_A F(x) dD(x)$ is well-defined, but in general it will be tricky to actually compute anything with this integral except in very specialized circumstances.
Now, while this is formally an answer, it is inadequate in a couple of respects, the first being that it totally sidesteps the specific computational question you raise, and the second being that if you don't know about amenable groups or measure theory, it is not likely to be terribly comprehensible. So let me try and address more directly what you've written.
It is correct, once we extend the density function as described above, that $A_0$ has density zero with respect to any Folner sequence. This follows from translation invariance - if we pick infinitely many distinct numbers $q_n$ strictly between 0 and 2, then the sets $A_0+q_n$ (i.e. the translations of $A_0$ by $q_n$) will all be mutually disjoint. In particular there is a finite number $N$ such that $D(\bigcup_{n=1}^N (A_0+q_n))>1$, which contradicts the fact that D assigns probability 1 to the entire space.
However, there are certainly going to be subsets of the space that have density which is strictly in between 0 and 1. One way to do this is to design a set $X$ so that for every $n$, $|X\cap F_n|/|F_n|$ is roughly 1/2.
As to your question about whether you can evaluate the integrals with respect to density in the way that you do, I will just say that it is very important to look carefully at how this "integral" has been defined, and not just assume that it will behave "similarly" to the integral on the real line.