Let $k>1$ be an element of $A$. Also let $f_k(x)= \left\lfloor \frac{x}{k} \right\rfloor$. Clearly we have $\left\lfloor \frac{k}{k} \right\rfloor = 1 \in A$. Note that for any positive integers $a,b$, we have $\left\lfloor \frac{a}{k^b} \right\rfloor = f_k^b(a)$. For any positive integer $n$, there exists an element $x$ of $A$ such that $x>k^{2n}+k^n$. Then $y=f_k^n(x) \in A$, and $\frac{x}{k^n} -1 < y \leq \frac{x}{k^n}$, so $k^n \leq \frac{x}{y} < \frac{x}{\frac{x}{k^n} -1} = \frac{xk^n}{x-k^n} = k^n+\frac{k^{2n}}{x-k^n} < k^n+1$, from which it follows that $\left\lfloor \frac{x}{y} \right\rfloor = k^n \in A$. Thus, for any nonnegative integer $n$, $k^n \in A$. Let $S_m=\{ m^n | n \in \mathbb{Z}^+ \}$ for any integer $m>1$. It follows that $A$ is a union of sets of the form $S_k$. We claim that if $k^a, k^b \in A$ for some positive integers $k, a,b$, and $gcd(a,b)=1$, then $k \in A$. By Bezout's lemma, there exists positive integers $x,y$ such that $ax-by=1$. So $\left\lfloor \frac{(k^a)^x}{(k^b)^y} \right\rfloor = k \in A$. We claim that if distinct $a,b \in A$, such that neither is a rational power of the other, then $A=\mathbb{Z}^+$. It suffices to show that any integer $z$ can be expressed in the form $z= \left\lfloor \frac{a^s}{b^t} \right\rfloor$ for some positive integers $s,t$. $z= \left\lfloor \frac{a^s}{b^t} \right\rfloor \iff z \leq \frac{a^s}{b^t} < z+1 \iff \ln z \leq s \ln a - t \ln b < \ln{(z+1)}$. Since $a$ and $b$ are not rational powers of each other, $\frac{\ln a}{\ln b}$ is irrational, so we may choose $x$ and $y$ such that $x \ln a - y \ln b $ is arbitarily small. Scaling up by a suitable factor, we find $s$ and $t$ such that $\ln z \leq s \ln a - t \ln b < \ln{(z+1)}$, thus $z= \left\lfloor \frac{a^s}{b^t} \right\rfloor$. It follows that $A = \mathbb{Z}^+$, or $A=S_k$ for some positive integer $k>1$. It is easy to check that both work.

Answer: $A=\mathbb{Z}^+$ or $A=\left\{ r^n | n \in \mathbb{Z}_{\geq0}\right\}$ for some integer $r \geq 2$ Clearly $1 \in A$. Denote by $r$ the minimal element of $A \setminus \left\{1\right\}$, and let $S_k = A \cap [r^k, r^{k+1})$ for all $k \geq0$. Claim 1: $S_k$ is non-empty for all $k \geq0$. proof: Assuming the contrary, there exists $k$ such that $S_k = \phi$ and $S_{k+1} \neq\phi$ because the number of non-empty sets in $\{S_k\}_{k=0}^{\infty}$ is infinite. $k \geq2$ because $S_0 = \left\{1 \right\}$ and $r \in S_1$. Take any $x \in S_{k+1}$. Then $\lfloor x/r \rfloor \in [r^{k}, r^{k+1}) $ should be an element of $S_k$ , contradiction. Now denote by $m(k)$ the minimal element of $S_k$. Claim 2: For any $a, b \in \mathbb{Z}_{\geq0}$, $m(a+b) \geq m(a)m(b)$. proof: For any $k \geq 0$ , $\lfloor m(k+1)/r \rfloor \in S_k$ so $\lfloor m(k+1)/r \rfloor \geq m(k)$ and $m(k+1) \geq r\cdot m(k)$. Multiplying this inequality for $k = a, a+1, ..., a+b-1$ gives $m(a+b) \geq r^{b}m(a)$. $m(a+b)/m(a)<r^{a+b+1}/r^a=r^{b+1}$ by definition, so $\lfloor m(a+b)/m(a) \rfloor \in [r^b, r^{b+1}), \lfloor m(a+b)/m(a) \rfloor \in S_b$ and the claim follows from $\lfloor m(a+b)/m(a) \rfloor \geq m(b)$. Claim 3: $m(k)=r^k$ for all $k \geq0$. proof: If there is $k$ such that $m(k) > r^k$, repeated application of Claim 2 shows $m(tk) \geq \{m(k)\}^t \geq (r^k + 1)^t $ for all integer $t \geq 1$. Hence $(r^k + 1)^t < r^{tk+1}$ or $(1+r^{-k})^t < r$ which is clearly false for sufficiently large $t$ . Therefore $\left\{ r^n | n \in \mathbb{Z}_{\geq0}\right\} \subseteq A$, and the condition holds when $A$ contains only integer powers of $r$. Suppose $A$ has an element $x$ that is not an integer power of $r$ and take minimal such $x$. Claim 4: For any non-negative integers $n, m$, $\lfloor r^n/x^m\rfloor \in \{0\} \cup A$ . proof: The proof goes on by induction on $m$ . The base case $m=0$ is true by Claim 3. Assume $m\geq1$ and that the claim is true for $m-1$. When $r^n \geq x^m$ , $\lfloor r^n/x^{m-1}\rfloor \geq x$ and $x$ are in $A$, so $\lfloor r^n/x^m \rfloor=\lfloor \frac{\lfloor r^n/x^{m-1} \rfloor}{x} \rfloor\in A$ as desired. Claim 5: $A = \mathbb{Z}^+$ . Let $\alpha = \log_r x$ . If $\alpha=q/p\in\mathbb{Q}$ where $p\geq2, q$ are relatively prime integers, $r^{1/p}$ is integer and $\lfloor x/r\rfloor=r^{\alpha-1} \in A$ in contradiction to minimality of $x$. Hence $\alpha$ is irrational, and the set $T := \{n\alpha - \lfloor n\alpha\rfloor : n\in\mathbb{Z}^+\}$ is dense on $(0, 1)$ . For any positive integer $N$, take $m \in \mathbb{Z}^+$ such that $r^{m-1}\leq N<r^m$ and $k\alpha-\lfloor k\alpha\rfloor\in T(k\in \mathbb{Z}^+)$ satisfying $\log_r(\frac{r^m}{N+1}) < k\alpha-\lfloor k\alpha\rfloor < \log_r(\frac{r^m}{N})$ . Now setting $m=k, n=\lfloor k\alpha\rfloor +m$ in Claim 4 gives $\lfloor\frac {r^{\lfloor k\alpha\rfloor}}{x^k}\cdot r^m\rfloor=\lfloor r^{\lfloor k\alpha\rfloor - k\alpha}\cdot r^m\rfloor=N\in A$ . $A = \mathbb{Z}^+$ clearly satisfies the problem condition, so in conclusion either $A = \left\{ r^n | n \in \mathbb{Z}_{\geq0}\right\}$ or $A = \mathbb{Z}^+$ .