Is following true?


For every given $c>0$ there is an $n\_c>0$ such that for every $n>n\_c$ there are integers $n<a,b<2n$ such that there are two positive integers $\frac{n}{2(\log n)^c}\approx\alpha<\beta\approx2\alpha$ such that
for every $x\in[\alpha,\beta]$ and $y\in[\alpha,\beta]$ we have
$$ax+by$$ always represents a prime or a number that does not have more than $c'(\log\log n)$ factors (including composites) for some $c'>0$.



Linear forms that avoid numbers with lot of factors
There is no such $c>0$. Indeed, assume that $n$ is large and $a,b,\alpha,\beta$ have the given property. Put $k:=\log n/\log\log n$. The number of prime factors of $a$ is at most $k+o(k)$, hence up to $(3/2)\log n$ there are at least $k/2-o(k)$ primes that don't divide $a$. So pick any $\lfloor k/3\rfloor$ such primes, and denote by $m$ their product. Then $m < n^{1/2} < \frac{n}{3(\log n)^c}$, and also $(a,m)=1$. Now it is easy to see that your numbers $ax+by$ contain a complete set of residues mod $m$, even if $y$ is an arbitrary fixed value. In particular, there is $x,y\in[\alpha,\beta]$ such that $ax+by$ is divisible by $m$, but then the number of divisors of this sum is at least $\tau(m)=2^{\lfloor k/3\rfloor}$. This number is much larger than $\log\log n$, a contradiction.


**P.S.** By a slight modification of the argument, one can improve the exponent $k/3$ to $k-o(k)$.


