Goldbach conjecture and other problems in additive combinatorics

It seems what you are asking is "If we have a precise asymptotic for the number of elements of a set, can we solve binary additive problems involving that set?"

The answer in general seems to be `no'. Let's consider Goldbach's conjecture that every large integer $n$ is the sum of two primes. It is not hard to see from pigeonholing that the typical $n$ will have at most $O( n / \log^2 n)$ solutions to $n=p+q$ within the primes. In fact, classical sieve theory easily establishes a uniform upper bound of this form unconditionally.

Now pick a rapidly increasing sequences of numbers $n'$ and remove from the set of primes those primes arising in solutions to $n'=p+q$ for that given $n'$. For each $n'$ we have removed at most $O(n' / \log^2 n')$ elements from the full set of primes, and so the asymptotic of the counting function of our set has not changed, however the assertion that every large integer is the sum of two elements from our modified set is now false.

You might object that my modified set of primes will not satisfy the more precise asymptotics (with error terms) that hold for the primes, such as the consequences of the (Generalized) Riemann Hypothesis or the Elliott-Halberstam conjectures. And this is true. However, there has been a lot of effort put into trying to deduce solutions to additive problems conditional on these conjectures, and even assuming these conjectures there is no known proof of either of the two famous additive problems (Goldbach and twin primes). Indeed there is an obstruction related to the "parity problem" in sieve theory which also enters the picture.

This does give rise to the following interesting question, which I do not know the answer to:

Does there exist a set of integers that satisfies the asymptotic behavior of the primes in arithmetic progressions (with the error term implied by GRH), but which fails to satisfy weak Goldbach?

A negative answer to this question would fairly conclusively yield a negative answer to your question.