Given $g(n) = \omega(f(n))$, then $g(n) + f(n) = \Omega(g(n))$?
There is a simpler proof. Assuming (as is usually the case) that $f$ is non-negative, we get $g(n) + f(n) \geq g(n)$, which implies that $g(n) + f(n) = \Omega(g(n))$. There is no need to assume anything.

If we know that $g(n) = \Omega(f(n))$, we can conclude further than $g(n) + f(n) = \Theta(g(n))$. And if $g(n) = \omega(f(n))$ then $g(n) + f(n) \sim g(n)$.