Clearly $\alpha>1$. By the well-known Theorem of Beatty, the complement of $A(\alpha)$ is just $A(\beta)$ where $\beta>1$ is such that $\frac{1}{\alpha}+\frac{1}{\beta}=1$. Now it is clear from a basic estimate on how many numbers there are in such a sequence that $\beta=2021$ (which of course indeed works) and hence $\alpha=\frac{2021}{2020}$ is the only choice.

@Tintarn Shouldn't both $\alpha$ and $\beta$ be irracional so you can use Beattyś theorem?

Aha, you are right of course! In any case, the answer $\alpha=\frac{2021}{2020}$ stands as the only choice, with a new (and much simpler) argument as follows: We can just directly count: The set $A(\alpha)$ has density $\frac{1}{\alpha}$ in the positive integers, while its complement by assumption has density $\frac{2020}{2021}$. Hence immediately $\alpha=\frac{2021}{2020}$ is the only choice and we only need to check that it indeed works.

If $\lfloor \alpha \rfloor>1\Rightarrow \lfloor 2\alpha\rfloor > 2 \Rightarrow 1, 2 \notin A(\alpha)$, so $\lfloor \alpha \rfloor=1 $ and we have $\lfloor 2\alpha \rfloor=2, \lfloor 3\alpha \rfloor=3, \dots , \lfloor (r-1)\alpha \rfloor=r-1, \lfloor r\alpha \rfloor =r+1, \dots, \lfloor (2021+r-1)\alpha \rfloor= 2022+r$, $ \dots \Rightarrow \lfloor (2021k+r)\alpha \rfloor=2021k+r+1 \Rightarrow \alpha\geq \dfrac{2021k+r+1}{2020k+r}$. Analogously we have that $ \lfloor (2021k+r-1)\alpha \rfloor=2021k+r-1\Rightarrow 2021k+r>(2021k+r-1)\alpha\geq 2021k+r-1\Rightarrow \alpha < \dfrac{2021k+r}{2020k+r-1}$. Hence we have to $$\dfrac{2021k+r+1}{2020k+r} \leq \alpha<\dfrac{2021k+r}{2020k+r-1}$$ So, like $\lim_{k\to\infty} \dfrac{2021k+r+1}{2020k+r}=\dfrac{2021}{2020}=\lim_{k\to\infty} \dfrac{2021k +r}{2020k+r-1}$ by the sandwich theorem $\alpha=\dfrac{2021}{2020}$.

Solution 2:= As previously proved $\lfloor \alpha \rfloor=1$, we now have two cases 1) $\alpha \in \mathbb{I}$ If $\alpha$ is irrational, beatty's theorem gives us that there exists $\beta$ such that $\dfrac{1}{\alpha}+\dfrac{1}{\beta}=1$ and all integers other than are in $A(\alpha)$ are in $A(\beta)$, i.e. $\lfloor \beta \rfloor,\lfloor 2\beta\rfloor, \lfloor 3\beta\rfloor,\dots$ leave all the same remainder when dividing by 2021, that is, $A(\beta)={r, 2021+r, 2021\cdot 2+r, \dots, 2021k+r, \dots }$. If $r>0 \Rightarrow \lfloor \beta \rfloor =r \Rightarrow \lfloor 2\beta \rfloor= 2r$ or $2r+1$, if $\lfloor 2\beta \rfloor=2r=r+2021\Rightarrow r=0$, absurd, if $ \lfloor 2\beta \rfloor=2r+1$ we will have $r=2020$ then $\lfloor 2019\alpha\rfloor=2019 \Rightarrow 2020+\alpha>2020\alpha\geq 2019+\alpha$ , but as $\lfloor 2020\beta \rfloor=2021$ we have $\alpha=2\Rightarrow r=0$, but if $r=0$ similarly we arrive at an absurd, so no have a solution in this case. 2) $\alpha \in \mathbb{Q}$ If $\alpha=\dfrac{a}{b}$ with $a, b \in \mathbb{N}$ and $gcd(a,2021)=1$ there is $k$ such that $ak\equiv 1(2021)\Rightarrow a( ck)\equiv c(2021)$ for all $c$, hence taking $n=ackb\Rightarrow \lfloor n\alpha \rfloor\equiv c(2021)$, absurd, thus $gcd(a, 2021)> 1$ which gives us 2 cases ($\alpha =1$ clearly gives us nonsense): 2.1) $\alpha=\dfrac{47}{b}$ (similar to $\alpha=\dfrac{47v}{b}$ or $\alpha=\dfrac{43r}{b}$) Suppose there are $k, j$ such that $\lfloor \dfrac{47k}{b} \rfloor=47j+r$, hence $k=bl+p$ with $p<b$ we have $\lfloor \dfrac{47p}{b} \rfloor<47$ but since $\lfloor \dfrac{47k}{b}\rfloor=\lfloor \dfrac{47p}{b}+47l \rfloor$ we have to $l=j $ then $ \lfloor \dfrac{47k}{b}\rfloor=\lfloor \dfrac{(bj+p)47}{b}\rfloor =\lfloor \dfrac{47p}{b}+47j\rfloor$ de so that $\lfloor \dfrac{47p}{b}=r$, nonsense, so an integer $x$ belongs to $A(\alpha)$ if and only if $x\equiv r(47)$ but hence, taking $x$ such that $x\equiv r(47)$ and $x\neq r(2021)$ we have that $x$ will also not be in $A(\alpha)$ but $x$ will not be $ r$ mod 2021, absurd. 2.2)$\alpha = \dfrac{2021}{b}$ (similar to $\alpha=\dfrac{2021s}{b}$), now we have that $\lfloor (b-1)\dfrac{2021}{b}\rfloor =2021 +\lfloor \dfrac{-2021}{b }\rfloor$ but we know that $\lfloor \dfrac{2021}{b}\rfloor=1$ and $\alpha>1$ so that $\lfloor \dfrac{-2021}{b}\rfloor=- 2\Rightarrow \lfloor (b-1)\dfrac{2021}{b}\rfloor=2021$, but since $\lfloor \dfrac{2021b}{b}\rfloor=2019$ we have $r=2020$ (because this was the only one skipped) hence $b\leq r$ Suppose there is $f$ such that $\lfloor \dfrac{2021f}{b} \rfloor=2021w+b$ (which is the same as assuming $b<r=2020$, hence $f=bo+x$ with $x<b \Rightarrow \lfloor \dfrac{2021f}{b} \rfloor=2021o+\lfloor \dfrac{2021x}{b} \rfloor$ but as $x<b$ we have that $\dfrac{2021x}{ b}<2021 \Rightarrow w=o$ and then $\lfloor \dfrac{2021x}{b} \rfloor=b$ absurd because as r=2020 for all $x<r$ we have that $\lfloor \dfrac{ 2021x}{b} \rfloor=x$ so we arrive at an absurdity and $f$ doesn't exist, so $b=r=2020$ and the only solution is $\alpha=\dfrac{2021}{2020}$.