Notice that two roots of product $p=1$ and sum $s$ verify the equation $x^2-sx+p=0$
$x^3=x^2x=(sx-1)x=sx^2-x=s(sx-1)-x=(s^2-1)x-s$
You can prove by induction that $x^n=f_n(s)x+g_n(s)$ with $\begin{cases}f_{n+1}(s)=sf_n(s)+g_n(s)\\g_{n+1}(s)=-f_n(s)\end{cases}$
For for these two roots $x,y$ you get $\begin{cases}x^{2020}=f_{2020}(s)x+g_{2020}(s)=ax+b\\y^{2020}=f_{2020}(s)y+g_{2020}(s)=ay+b\end{cases}$
By difference and since $x-y\neq 0$ as the problem specifies explicitely two distinct real numbers
we get: $\begin{cases}a=f_{2020}(s)\\b=g_{2020}(s)\end{cases}$
As long as $s$ is an integer, it follows that $f_n(s)$ and $g_n(s)$ are also integers.
Therefore you can generate $(a,b)$ integers and $(x,y)$ real roots of product $1$ that answer the problem by starting from $x^2-sx+1=0$ and varying $s\in\mathbb Z$.
- We still need to prove injectivity to get that they are in infinite quantity.
But notice that $f_n(s)=\dfrac{x^n-y^n}{x-y}$ and that $xy=1\implies |x|>1\text{ and }|y|<1\quad$
(or vice-versa, WLOG let assume $|x|>|y|$)
It ensues that $|y|^{2020}\ll 1$ and that $a=\operatorname{round}(\frac{x^{2020}}{x-y})\ge\operatorname{round}(\frac 12x^{2019})$
And since $x=\frac{s+\sqrt{s^2-4}}{2}\nearrow$ when $s\nearrow$ we have also $a\nearrow$, and strict monotonicity gives injectivity.
Edit:
Note that for $s$ large, $x\sim s$ and $a\sim s^{n-1}$, so you are pretty much ensured of injectivity for any exponent instead of $2020$ as long as your $s$ is large enough, which is not an issue since $s$ can be taken arbitrarily in $\mathbb Z$.