Distinct sets of primes have distinct products, hence you are counting the number of square-free $Y$-smooth integers below $X$.

The number of $Y$-smooth integers below $X$ is $$\Psi(x,y)=x\rho(u)+O(x/\log y),$$ where $x=y^u$, and $\rho$ is Dickman’s function, which is $1-\log u$ for $u\in[1,2]$. In the range you are interested in (or more generally, if $X=Y^{1+o(1)}$), we have $u=1+o(1)$, hence $\rho(u)=1+o(1)$, and $$\Psi(X,Y)=X+o(X).$$ In other words, the number of integers below $X$ that are not $Y$-smooth is $o(X)$. Since there are $(6/\pi^2)X+O(\sqrt X)$ square-free integers below $X$, the number of your sets is $$\frac6{\pi^2}X+o(X).$$

Distinct sets of primes have distinct products, hence you are counting the number of square-free $Y$-smooth integers below $X$.

The number of $Y$-smooth integers below $X$ is $$\Psi(X,Y)=X\rho(u)+O(X/\log Y),$$ where $X=Y^u$, and $\rho$ is Dickman’s function, which is $1-\log u$ for $u\in[1,2]$. In the range you are interested in (or more generally, if $X=Y^{1+o(1)}$), we have $u=1+o(1)$, hence $\rho(u)=1+o(1)$, and $$\Psi(X,Y)=X+o(X).$$ In other words, the number of integers below $X$ that are not $Y$-smooth is $o(X)$. Since there are $(6/\pi^2)X+O(\sqrt X)$ square-free integers below $X$, the number of your sets is $$\frac6{\pi^2}X+o(X).$$