Meaning of "for sufficiently large" in infinite products

This is just another way of saying that for all but finitely many primes $p$, we have $\alpha(p) = 0$. In particular, the "largeness of the prime" depends upon the $a$ you are given to start with.

Specifically how this relates to your case. You are given $a \geq 1$, then, for the most unrefined bound, $p > a$ can never be factors of $a$.

What that means is that, for each $n>1$, you can write $n$ as$$2^{\alpha(2)}\times3^{\alpha(3)}\times5^{\alpha(5)}\times7^{\alpha(7)}\times\cdots,$$where $\alpha(p)=0$ if $p$ is sufficiently large; in other words, you only have $\alpha(p)\ne0$ for finitely many primes.

Note that here “sufficiently large” depends on $n$. For $n=10$, “sufficiently large” means $p>5$, whereas for $n=74$, it means $p>37$.

Does that mean large primes can never be factors of any integers?

Of course not: each prime is a factor of itself.

What's sufficiently large depends on $a$. it means for each $a$, there is some $N$ for which any $p>N$ is sufficiently large. We can take $N$ to be $a$'s greatest prime factor.