I will give a complete answer when $1,1/r$ and $1/s$ are linearly independent over $\mathbb Q$ and a recipe to compute your $t$ otherwise.
First of all, notice that $n$ lies in a Beatty sequence $\lfloor m\alpha\rfloor$, where $\alpha>1$ if and only if $\{n/\alpha\}>1-\frac{1}{\alpha}$. Indeed, if $n=\lfloor m\alpha \rfloor$, then
$$
\frac{n}{\alpha}=\frac{m\alpha-\{m\alpha\}}{\alpha}=m-\frac{\{m\alpha\}}{\alpha}, \text{ so }\{n/\alpha\}>1-\frac{1}{\alpha}.
$$
On the other hand, if $\{n/\alpha\}>1-\frac{1}{\alpha}$, then one can set $m=\lfloor n/\alpha \rfloor+1=n/\alpha+1-\{n/\alpha\}$ and get
$$
m\alpha=n+\alpha\left(1-\{n/\alpha\}\right),
$$
so $\lfloor m\alpha \rfloor=n$, as needed.
Now, Case 1: $1,1/r$ and $1/s$ are linearly independent over $\mathbb Q$. In this case, it is known (say, by Weyl's criterion), that the points
$$
t_n=(n/r \mod 1, n/s \mod 1)\in \mathbb R^2 \diagup \mathbb Z^2
$$
are uniformly distributed on the torus. This means that for $N\to +\infty$ the proportion of $n\leq N$ with
$$
\{n/r\}>1-\frac{1}{r}\text{ and }\{n/s\}>1-\frac{1}{s}
$$
is equal to the measure of the set $(x,y) \in \mathbb R^2\diagup \mathbb Z^2$ with $x>1-\frac{1}{r}, y>1-\frac{1}{s}$, hence the inverse $t$ of this limiting proportion is equal to $rs$. For example, for $\pi$ and $e$ we get $\pi e\approx 8.5397$ and for $\sqrt{3}$ and $\sqrt{5}$ we get $\sqrt{15}\approx 3.873$.
Case 1.5 is when both $r$ and $s$ are rational. In this case, your problem reduces to enumeration of certain congruence classes $\mod \mathrm{num}(r)\mathrm{num}(s)$.
Finally, Case 2 is when $1/r$ is irrational, but for some $a,p,q\in \mathbb Z$ we have $1/s=p/q\cdot1/r+a/q$. Without loss of generality one can assume that $p,q>0$, $a\geq 0$. Next, let us divide $\mathbb N$ into congruence classes $\mod q$ and evaluate the (relative) density $\delta_b$ of all $n\equiv b\pmod q$ such that
$$
\{n/r\}>1-\frac{1}{r}\text{ and }\{n/s\}>1-\frac{1}{s}.
$$
In this case, the answer would be
$$
t=\frac{q}{\delta_0+\delta_1+\ldots+\delta_{q-1}}.
$$
Fix $b$. The congruence $n\equiv b\pmod q$ means that $n=qm+b$. Now, we need
$$
\{qm/r+b/r\}>1-\frac{1}{r}\text{ and }\{pm/r+b/s\}>1-1/s.
$$
Since $1/r$ is irrational, $\{m/r\}$ is equidistributed $\mod 1$. This means that our relative density $\delta_b$ is equal to the measure of all $x\in [0,1]$ such that
$$
\{qx+b/r\}>1-\frac{1}{r}\text{ and }\{px+b/s\}>1-1/s.
$$
(By relative density I mean the density of corresponding $n$ in the residue class, i.e. the actual density of your $n$ with $n\equiv b\pmod q$ is $\delta_b/q$)
Best Answer
Define $a_n=\Phi_n(1)$ except $a_1=1$. These values are uniquely determined by $$n=\prod_{d \mid n}a_d.$$ From this it would be quick to get $a_n=c_n$ but , as requested, we won't. An important step is to observe that $\gcd(a_m,a_n)=1$ if $\gcd(m,n)=c \lt m \lt n.$ This follows from $\gcd(\frac{m}{c},\frac{n}{c})=1$ by writing $m,n,c$ as products of the $a_i$ and simplifying.
If follows that $$n!=\prod_{k=1}^na_k^{\lfloor n/k\rfloor}$$ and $$\binom{n}{m}=\frac{n!}{m!(n-m)!}=\prod_{k \le n}a_k^{e_{n,m,k}}$$ where the exponent $$e_{n,m,k}=\lfloor n/k \rfloor-\lfloor m/k \rfloor-\lfloor (n-m)/k \rfloor$$ Note that $0 \le e_{n,m,k} \le 1$ and $e_{n,m,k}=0$ if $k \mid m$ and $k \mid n.$
We hope to show that $$\gcd{\Large (}\binom{n}{1},\binom{n}{2},\cdots,\binom{n}{n-1}{\Large )}=a_n.$$ We certainly know that $a_n$ divides each of these binomial coefficients. We will now see that it is sufficient to consider just $\binom{n}{1}$ and the various $\binom{n}{n/p}$ as $p$ ranges over the prime divisors of $n.$
Let $n=q_1q_2\cdots q_j$ where the $q_j=p_j^{e_j}$ are powers of distinct primes and define $G_0=\binom{n}{1}$ and $G_i=\gcd(G_{i-1},\binom{n}{n/p_i})$ for $1 \le i \le j.$ Then $G_i$ is the product of the $a_k$ with $q_1q_2\cdots q_i \mid k \mid n.$ and $G_j=a_n$
I will illustrate with the example of $n=900=2^23^25^2$. So $G_0=\binom{900}{1}$, $G_1=\gcd(G_0,\binom{900}{900/2})$, $G_2=\gcd(G1,\binom{900}{900/3})$ and $G_3=\gcd(G_2,\binom{900}{900/5}).$ I claim that $G_3=a_{900}.$
$G_0=\binom{900}{1}$ is the product of the $26$ $a_k$ with $1 \lt k \mid 900.$ Of these, the $9$ which are multiples of $2^2$, namely $a_4,a_{12},a_{20},a_{36},a_{60},a_{100},a_{180},a_{300}$ and $a_{900}$ also divide $\binom{900}{900/2}=\binom{900}{450}$, the other $17$ have $e_{900,450,k}=0.$ There are many other $a_k$ with $e_{900,450,k}=1$ but all of them are relatively prime to $G_0$ by the important step above. Hence $G_1$ is exactly the product of the $9$ $a_k$ with $4 \mid k \mid 900$. Now $G_2=\gcd(G_1,\binom{900}{300})=a_{36}a_{180}a_{900}$ because the only other $a_k$ dividing $\binom{900}{300}$ are such that $k$ neither divides nor is a multiple of $4,12,20,60,100,300$ and hence all are relatively prime to $G_1$ by the important step. Finally $G_3=\gcd(G_2,\binom{900}{180})=a_{900}$ because the only other $a_k$ with $e_{900,180,k}=1$ are relatively prime to both $a_{36}$ and $a_{180}$.
I believe this does what was requested.
It may be helpful to look at a more abstract setting where essentially the same proof goes through. The idea is to replace each integer $n$ by a generalized integer $I_n$, replace $a_n$ by some appropriate $A_n$, define a generalized factorial $[I_n]!=I_1I_2\cdots I_n$ and then a generalized binomial coefficient $${n \brack m}=\frac{[I_n]!}{[I_m]![I_{n-m}]!},$$ and deduce that $\gcd\left( {n \brack 1},{n \brack 2},\cdots,{n \brack {n-1}}\right)=A_n$. What is meant by $\gcd$ needs to specified in some cases.
Suppose that we have a commutative ring $\mathcal{R}$ and within it a sequence of elements $I_1,I_2,I_3,\cdots$ with the property
$$\gcd(I_m,I_n)=I_{\gcd(m,n)} \tag{**}$$
or perhaps merely the weaker property
$I_d \mid I_n$ when $d\mid n$ $(*)$
Based just on $(*)$ we have that there are elements $A_1,A_2,\cdots$ uniquely defined by $I_n=\prod_{d\mid n}A_d.$ (Although we will not need it, it then follows that $A_n=\prod_{d\mid n}I_n^{\mu(n/d)}$ where $\mu$ is the classic Mobius function.)
A familiar and obvious choice with (**) is $I_n=1+X+\cdots+X^{n-1}.$ Then we have the cyclotomic polynomials $A_n=\Phi_n(X)$ defined by $$I_n=\prod_{d\mid n}A_d.$$ (Except that $A_1=1$.)
Examples with (*) include
Asides: We will always be able, if desired, to assume $A_1=I_1=1$ by replacing $I_n$ with $\frac{I_n}{I_1}$ as in 4. Example 1 is the case $u=v=1$ of 4 while example 2 is the case $u,v=(1\pm \sqrt{5})/2$ with the convenient feature $uv=-1$. The obvious choice above is example 4 with $u=X,v=1.$
In any case, given just (*) we can define a generalized factorial $$[I_n]!=I_1I_2I_3\cdots I_n$$ and it follows that $$[I_n]!=\prod_{k=1}^nA_k^{\lfloor n/k\rfloor}$$ as in the integer case.
We can also define a generalized binomial coefficient by
$${n \brack m}=\frac{[I_n]!}{[I_m]![I_{n-m}]!}=\prod_{k \le n}A_k^{e_{n,m,k}}$$
The argument above becomes $\gcd\left( {n \brack 1},{n \brack 2},\cdots,{n \brack {n-1}}\right)=A_n$ provided we do have the stronger condition $(**)$. This requires knowng what is meant by $\gcd$ and that it behaves as expected. I don't think (**) holds for example 4, but it does when $v=1$. The fact that it hold in the case of example 2 is perhaps due to the extra relation $uv=-1$.
In a ring $\mathcal{R}$ I will take as the definition of $\gcd(U,V)=W$ ($W$ is a gcd of $U$ and $V$) to be
$W \mid U$ and $W \mid V$ and $US+VT=W$ for some $S,T \in \mathcal{R}$.
Note that we do not assume that every pair $U,V$ have a $\gcd$. In $\mathbb{Z}[X]$ We do not have a $\gcd$ for $U=\Phi_4=X^2+1$ and $V=\Phi_2=X+1$. We can get $US+VT=2$ but not $1$.
Actually finding the the explicit cofactors might be impractical.
The fact that they exist follows (in certain of the examples above) from
but even in the first case we don't have a simple way to explicitly find $s,t$ with $ns+mt=\gcd(n,m)$, even in the case that the right-hand side is $1.$ Furthermore, the claim that this $\gcd$ behaves appropriately depends on iterated application of facts such as:
Given (in some ring) that $\gcd(U,V)=\gcd(U,V')=1$ in the sense that there are $S,T,S',T'$ with $US+TV=1$ and $U'S'+T'V'=1$, it follows that $\gcd(U,VV')=1$ since $$U{\LARGE(}USS'+ST'V'+S'TV{\LARGE)}+VV'(TT')=1.$$
It is something I have asked about before, although not very clearly.