What you write as a condition for $f\in\Omega(g)$ isn't true. $f\in\Omega(g)$ if $f(n)\geq kg(n)$ for some $k$ and big enough $n$, so certainly for instance $n^3\in\Omega(n^2)$, but the limit of $n^3/n^2$ is $\infty$. I think you want $0<\lim f/g\leq\infty$.
With that change, this goes quickly: just write $\lim_{n\to\infty} \frac{f(n)+g(n)}{f(n)}=\lim_{n\to\infty}1+\frac{g(n)}{f(n)}>0$ To be precise, we might have the last limit be nonzero by not existing, rather than by being positive, as for instance if $f(n)=\langle 1,2,1,2,1...\rangle$ and $g(n)=\langle 2,1,2,1,...\rangle$. This would give $f+g=3$, and $3/f$ oscillates between $3$ and $3/2$ without converging. So your statement should really read $f+g\in\Omega(f)$ provided $\lim_{n\to\infty} (f+g)/f$ exists.