Proof that $G$ is a yes-instance of IDS $\iff$ $f(G)$ is a yes-instance of SAT
Yes, your proof of $\Rightarrow$ is correct, but note that you still have to prove the second direction to complete the reduction.            


  Now we know that IDS is NP-hard        


Unfortunately, we don't. We'd need a reduction from SAT to IDS and not the other way around.
To convince yourself of why it makes sense, a reduction from $A$ to $B$ means that if we had an algorithm that solves $B$, we could use it to trivially solve $A$. So, in your case - if we had an algorithm that solves SAT we would be able to solve IDS, but that's not what we want to show. We want to claim that if we could solve IDS then we could solve SAT, which means IDS is "at least as hard as SAT".         

Also, notice that you need a karp reduction and not just a  many-one reduction. That is, a many-one reduction which can be computed in polynomial time (your reduction achieves that)


  Furthermore we have to provide an efficient algorithm which uses non-determinism to show that IDS is a member of NP, correct?


Correct.
Assuming we have provided a karp reduction from SAT to IDS, or equivalently - we proved that $\text{SAT}\le_p \text{IDS}$, then all we know is that IDS is NP-hard, and we need to prove that $\text{IDS} \in \text{NP}$ to get that IDS is NP-complete.