Your numbers have positive lower density. To see this let $z$ be a positive integer to be fixed later, and denote $$c:=\prod_{p \leq z}(1-1/p).$$ Consider all square-free integers $x < n \leq 2x$ which are composed of primes $z < p \leq \sqrt{x}$. Note that these numbers satisfy the requirements. Their number, by a crude estimate, is at least $$cx+O(1)-\sum_{\sqrt{x} < p \leq 2x}(cx/p+O(1)) - \sum_{p>z}x/p^2 sum_{z < p \geq (c-c\log leq \sqrt{2x}}(cx/p^2 +O(1)),$$ which is at least $$c(1-\log 2-1/z+o(1))x.$$ That is, we have positive the lower density once is at least $$c(1-\log 2) > 1/z. 2-1/z)/2.$$ For $The z:=5$ the left hand side is exceeds $\gg 1/\log (z)$, therefore the inequality holds 0.0142$, while for$z$sufficiently large. Numerical calculation shows that$z:=23$does the job, this choice proves that the lower density z:=17$ it exceeds 0.003$0.0223$.
Your numbers have positive lower density. To see this let $z$ be a positive integer to be fixed later, and denote $$c:=\prod_{p \leq z}(1-1/p).$$ Consider all square-free integers $x < n \leq 2x$ which are composed of primes $z < p \leq \sqrt{x}$. Note that these numbers satisfy the requirements. Their number, by a crude estimate, is at least $$cx+O(1)-\sum_{\sqrt{x} < p \leq 2x}(cx/p+O(1)) - \sum_{p>z}x/p^2 \geq (c-c\log 2-1/z+o(1))x.$$ That is, we have positive lower density once $$c(1-\log 2) > 1/z.$$ The left hand side is $\gg 1/\log (z)$, therefore the inequality holds for $z$ sufficiently large. Numerical calculation shows that $z:=23$ does the job, this choice proves that the lower density exceeds 0.003.