Construction for $r\geq s$ Firstly, note that all pairs $(r, s)$ with $r\geq s$ work because we can write \[a_{n+2} = \prod\limits_{i=1}^{n} \left(1+\frac{a_{i+1}^r}{a_{i}^s}\right)=\left(1+\frac{a_{n+1}^r}{a_n^s}\right) \prod\limits_{i=1}^{n-1} \left(1+\frac{a_{i+1}^r}{a_{i}^s}\right)=a_{n+1} \left(1+\frac{a_{n+1}^r}{a_n^s}\right).\]Therefore, if we pick $a_1=a_2=1$, then $a_3=2$, and by induction since $a_n\mid a_{n+1}$ and $r\geq s$, we get $a_n^s\mid a_{n+1}^r$, so $1+\frac{a_{n+1}^r}{a_n^s}\in \mathbb{N}$. This implies that $a_{n+2}$ is an integer, and moreover, $a_{n+1}\mid a_{n+2}$, so we're done by induction. All sequence members have the same set of prime divisors Now, we focus on the more interesting case $r<s$. Assume that there exists such a $p$, for which $p\mid a_k$, but $p\nmid a_{k-1}$ for $k\geq 2$. If $k=2$, then $\nu_p(a_3) = 0\Longrightarrow\nu_p(a_4) < 0$, which is impossible, so assume that $k\geq 3$, in which case: \[\nu_p(a_{k+1}) = \nu_p(a_k) + \nu_p\left(1+\frac{a_{k}^r}{a_{k-1}^s}\right) = \nu_p(a_k) > 0.\]But now $a_{k+2} = a_{k+1}\left(1+\frac{a_{k+1}^r}{a_{k}^s}\right)$ and $\nu_p(a_{k+1}) = \nu_p(a_k) > 0$, so $\nu_p(a_{k+2}) < \nu_p(a_{k+1})$ because $r<s$, and by induction we get that for all $i\geq 1$ we have $\nu_p(a_{k+i+1}) < \nu_p(a_{k+i})$, which is impossible, as we're working with positive integers. Therefore, if $p\mid a_k$, then by downward induction, we get that $p\mid a_1$. Also, if $p\mid a_k$ and $p\nmid a_{k+1}$, then $\nu_p(a_{k+2})<0$, which is impossible. So, every sequence member must have the same set of prime divisors. Bounding of the $\nu_p$'s and finishing Notice that as $r<s$, and $a_{n+2} = a_{n+1}+\frac{a_{n+1}^{r+1}}{a_{n}^s}$, we must have $(r+1)\nu_p(a_{n+1})>s\nu_p(a_n)$ for any prime divisor of the sequence as $p\mid a_{n+2}$ and $p\mid a_{n+1}$, so $\nu_p\left(a_{n+1}^{r+1}/a_{n}^s\right)>0$. However, as $r+1\leq s$, this gives $\nu_p(a_{n+1}) > \nu_p(a_n)$. The last inequality implies that: \[\nu_p(a_{n+2}) > \nu_p(a_{n+1})\Longrightarrow \nu_p(1+a_{n+1}^r/a_{n}^s)>0\Longrightarrow r\nu_p(a_{n+1}) = s\nu_p(a_n).\]As this holds for all primes $p$, then we can write $a_n^s = a_{n+1}^r$, which would result in $a_{n+2} = 2a_{n+1}$. This breaks the requirement $a_{n+2}^r = a_{n+1}^s$, so there are no solutions when $r<s$.

A funny problem cause you don't need to look deep to find the contradiction, $n=1,2,3$ are enough. Construction for $r \geq s$: Just take $a_1=a_2=1$. By induction $\frac{a_{n+1}^r}{a_n^s} \in \mathbb{Z_{+}}$. After that every factor in multiplying is a positive integer, thus we are done. Proof for $r<s$. $n=1,2$: $a_3=1+\frac{a_2^r}{a_1^s}$. $a_4=a_3(1+\frac{a_3^r}{a_2^s})=a_3+\frac{a_3^{r+1}}{a_2^s}$, from which $a_3^{r+1} \vdots a_2^s$. Suppose prime $p | a_2$, then $a_3 \vdots p$, then $\frac{a_2^{r+1}}{a_3^s}$ is not a multiple of $p$, hence $v_p(a_2^{r+1})=v_p(a_3^s)$. Because it is true for every such $p$, $a_2^{r+1}=a_3^s$. Then $a_3=2$. So $2^{r+1} \vdots a_2^s$, remember $s > r$. Then either $a_2=1$ or $a_2=2$. If latter occurs, then $r+1=s$ and $2^r=a_1^s$, a clear contradiction. Thus $a_1=a_2=1$. Now $a_4=2(2^r+1)$. Finally, $n=3$ gives us $a_4^{r+1} \vdots a_3^s$, from which we must have $r+1=s$. Calculating $a_5$ gives us $a_5=(2^r+1)(2+(2^r+1)^r)$, which is an odd number. But we should have $a_5^{r+1} \vdots a_4^s \vdots 2$, which is a contradiction.