Let's try induction.
The base case is easy (although not as trivial as you write in your question).
Now, assume any word in $L\cap \{0,1\}^n$ is generated by $G$. Let's take a word $w$ in $L\cap \{0,1\}^{n+1}$ and show it is generated by $G$.
Assume $w=0^a1^b$. We know that $b\ge 2(a-1)$. Note that $0^{a-1}1^{b-1}\in L\cap \{0,1\}^n$ (i.e., $b-1\ge 2(a-2)$) so it must be generated by $G$ . Since there is only one terminal derivation $S\to 0$, this must be the last derivation.
It follows that there must be a sequence of derivations in $G$ which looks like:
$$S \to^* 0^{a-2}S1^{b-1} \to 0^{a-1}1^{b-1}$$
with the last transition $S\to 0$. Here you need to explain why $0^{a-2}S1^{b-1}$ is the only possible sentinel phrase that yields $0^{a-1}1^{b-1}$ (and it can't be, e.g., of the form $0^{a_1}S0^{a_2}1^b$); This is a simple argument that I'll leave you to complete.
So if $S \to_{G}^* 0^{a-2}S1^{b-1}$ we can now take the second transition and then the third, and get
$$ S \to^* 0^{a-2}S1^{b-1} \to 0^{a-2}S11^{b-1} \to 0^{a-2}011^{b-1} = 0^a1^b$$
and we are done.
