EDIT: this question of mine has received little attention, perhaps in part because it was stated in a too general and complicated way. So let me give it a second chance:



> 
> Fix an integer $r \geq 0$. Let $E\_r(x)$ be the number of square free integers less than $x$ having exactly $r$ prime factors $p$ congruent to $2 \pmod{3}$ (and an arbitrary number of other prime factors). What is the order of magnitude of $E\_r(x)$ when $x$ goes to $\infty$ ?
> 
> 
> 


When $r=0$, one has $E\_0(x) \asymp x/\log^{1/2}(x)$ (see below). Even when $r=1$,
I don't know the answer, or a lower bound for $E\_1(x)$ that is better than this one for $E\_0(x)$. Thanks for any
idea or reference pointing to a method that can be used to solve this question.


ORIGINAL POST FOLLOWS:


I begin by quoting a theorem found in Serre's paper "divisibilité de certaines functions
arithmétiques", with a proof attributed to Raikov, Winter, Delange (theorème 2.4 - I have slightly changed and simplified the statement, as the $E$ here is the complement $E'$ of $E$ in Serre's paper.).


Let $E$ be a set of integers satisfying the following condition :


(MM) For $n,m$ relatively prime integers, $nm \in E$ if and only if $n \in E$ and $m \in E$, 


Let $P$ be the set of primes that are in $E$. Assume that $P$ has analytic density $\alpha > 0$. Let $E(x)$ the number of elements of $E$ less than $x$. Then when $x$ goes to infinity,
$$E(x) \sim c x /\log^\{1-\alpha}(x)$$ for some constant $c>0$. 


Example: Let $P$ be a set of primes of density $\alpha>0$, $E$ the set of square-free 
integers whose prime factors are all in $P$. Then $E$ satisfies the hypothesis (MM).
So does $F=$ the set of all integer who's prime factors are all in $P$. Both satisfies
the conclusion of the theorem, though with a different constant $c$.


Now let $P,E$ be as above, and let $r \geq 0$ be an integer. Let $E\_r$ be the set of integers that are products of an element of $E$ and exactly $r$ distinct prime not in $P$. My question is:


What is an order of magnitude of $E\_r(x)$ ?


The case $r=0$ is the theorem above, and I suppose the answer for $r>0$
does only depend on $P$, not on $E$, as for $r=0$. But in case it is not true, or in case that helps, the case of interest for me is the case where $E$ is as in the example.
So if I may reformulate my question in this special case: if $P$ is a set of prime of density $\alpha>0$, and $E$ is the set of square-free integers which are product
of $r$ primes not in $P$ and arbitrary number of primes in $P$, what is the order of magnitude of $E\_r(x)$?


A word on my motivation, which turns around the beautiful aforementioned paper of Serre.
Consider for example the modular form $\Delta$, modulo $3$. if $E$ is the set of $n$ such that the $n$-th coefficient of that form is non-zero, that is such that $3 \nmid \tau(n)$, then $E$ satisfies (MM), $P$ is the set of primes congruent to $1$ modulo $3$, and $E(x) \sim cx log^{1/2}(x)$.
This is in Serre, as $\Delta$ is an eigenform. Noe consider $\Delta^2$ modulo $3$ which is not an eigenform, and let $F$ the set of $n$-th coefficient of that form is non-zero. Then it is easy to see that $F$ contains $E\_1$ (and I think, is not too much bigger than $E\_1$). 



Density of a set of integers
Here's an attempt to say something about your nice question. It surely follows from Igor Rivin's nice sketch and the Math Overflow question he linked to. Fix $r$. The quantity $E\_r(x)$ in the edited question satisfies


$$
E\_r(x)\gg \sum\_{\substack{m \leq x \\ \omega(m)=r\\ \mu(m)^2=1\\p|m \Rightarrow p \equiv 2 \bmod 3}}\sum\_{\substack{n \leq x/m\\ \mu(n)^2=1\\p|n \Rightarrow p\equiv 1 \bmod 3}}1.
$$


The inner sum in the above, by what you wrote about $E\_0(x)$, satisfies


$$
\sum\_{\substack{n \leq x/m\\ \mu(n)^2=1\\p|n \Rightarrow p\equiv 1 \bmod 3}}1 \gg \frac{x}{m(\log x)^{1/2}}.
$$


By induction on $r$, say, and using $\sum\_{\substack{p \leq x\\p \equiv 2 \bmod 3}}\frac{1}{p}\gg \log\log x$, one has


$$
\sum\_{\substack{m \leq x \\ \omega(m)=r\\ \mu(m)^2=1\\p|m \Rightarrow p \equiv 2 \bmod 3}}\frac{1}{m}\gg (\log\log x)^r.
$$


So one has
$$
E\_r(x) \gg \frac{x(\log\log x)^r}{(\log x)^{1/2}}
$$
where the implied constant depends on $r$.


