We will prove that exactly all pairs $(a, b)$ with $\operatorname{gcd}(a, b)=1$ and with $a+b$ odd satisfy.

First, assume that $\operatorname{gcd}(a, b)=n \neq 1$. Consider the function

$$g(x)=\left\{\begin{array}{lll} x+1 & \text { if }\lfloor x\rfloor \not \equiv 0 & \bmod n \\ x+1-n & \text { if }\lfloor x\rfloor \equiv 0 & \bmod n \end{array}\right.$$

This function is not equal to $\operatorname{id}_{\mathbb{R}}$ because $n \neq 1$. Then $\lfloor g(x)\rfloor \equiv\lfloor x\rfloor+1 \bmod n$. Therefore, the numbers $\lfloor x\rfloor,\lfloor g(x)\rfloor,\lfloor g(g(x))\rfloor, \ldots,\left\lfloor g^{n-1}(x)\right\rfloor$ have all residue classes $\bmod n$. Thus $g^{n}(x)=x+\underbrace{1+\ldots+1}_{n \text { times }}-n=x$. By induction, it follows that $g^{c n}(x)=x$ for all natural $c$, so since $n \mid a, b$ it also holds that $g^{a}(x)=x$ and $g^{b}(x)=x$. Therefore,

$$g^{a}(x) g^{b}(y)+g^{b}(x) g^{a}(y)=x y+x y=2 x y$$

Thus the function $f=g \neq \operatorname{id}_{\mathbb{R}}$ satisfies the functional equation in this case.
Now consider the case where $a+b$ is even. Take the function $h(x)=-x$. Then by simple induction, $h^{c}(x)=(-1)^{c} x$. Therefore,

$$h^{a}(x) h^{b}(y)+h^{b}(x) h^{a}(y)=(-1)^{a+b} x y+(-1)^{a+b} x y=2 x y$$

Thus the function $f=h \neq \operatorname{id}_{\mathbb{R}}$ satisfies the functional equation.
Now assume that $\operatorname{gcd}(a, b)=1$ and that $a+b$ is odd. With $x=y$ we see

$$f^{a}(x) f^{b}(x)=x^{2}$$

If we multiply the functional equation by $f^{a}(x) f^{a}(y)$, we get

$$\begin{aligned} 2\left(x f^{a}(y)\right)\left(y f^{a}(x)\right) & =f^{a}(x) f^{b}(y) f^{a}(x) f^{a}(y)+f^{b}(x) f^{a}(y) f^{a}(x) f^{a}(y) \\ & =(\left(f^{a}(x)\right)^{2} \underbrace{f^{a}(y) f^{b}(y)}_{=y^{2}}+(\left(f^{a}(y)\right)^{2} \underbrace{f^{a}(x) f^{b}(x)}_{=x^{2}} \\ & =\left(y f^{a}(x)\right)^{2}+\left(x f^{a}(y)\right)^{2} \end{aligned}$$

This is the equality case of the arithmetic and geometric mean inequality. More beautifully, we can rewrite it as

$$\left(y f^{a}(x)-x f^{a}(y)\right)^{2}=0$$

Thus we conclude that $y f^{a}(x)=x f^{a}(y)$. With $y=1$ we see $f^{a}(x)=c_{1} x$ for some $c_{1} \in \mathbb{R}$. Similarly, multiplying by $f^{b}(x) f^{b}(y)$ gives $f^{b}(x)=c_{2} x$ for some $c_{2} \in \mathbb{R}$. If one of the constants were 0, then the left side of the functional equation would always be 0, but the right side would not. Therefore, both constants are not 0. Since $\operatorname{gcd}(a, b)=1$, there exist integers $p$ and $q$ such that $a p+b q=1$. Assume without loss of generality that $p$ is positive and $q$ is negative, and write $r=-q$. Then $a p=1+r b$ with $p$ and $r$ positive. We see

$$\begin{aligned} c_{1}^{p} x=\left(f^{a}\right)^{p}(x) & =f^{a p}(x) \\ & =f^{1+r b}(x) \\ & =f\left(\left(f^{b}\right)^{r}(x)\right) \\ & =f\left(c_{2}^{r} x\right) \end{aligned}$$

Thus $f(x)=\frac{c_{1}^{p}}{c_{2}^{r}} x=d x$. If we substitute this function, we see $2 d^{a+b} x y=2 x y$, so $d^{a+b}=1$. This implies $d=1$ because $a+b$ is odd. This means that $f(x)=x$ is the only function that potentially satisfies. It is easy to see that this function also actually satisfies, so all pairs $(a, b)$ of natural numbers for which $f(x)=x$ is the only function $f: \mathbb{R} \rightarrow \mathbb{R}$ that satisfies

$$f^{a}(x) f^{b}(y)+f^{b}(x) f^{a}(y)=2 x y$$

for all $x, y \in \mathbb{R}$ are exactly the pairs $(a, b)$ where $a+b$ is odd and $\operatorname{gcd}(a, b)=1$.