I have seen other proofs of this result using Burnside's lemma in G acts on X transitively, then there exists some element that does not have any fixed points
However, I tried proving it just using the orbit stabilizer theorem and want to check if my approach is correct. The proof goes as follows:
Let $a \in A$, then we know by the orbit-stabilizer theorem that $|\mathcal{O}(a)| = |G: Stab_G(a)|$. Since $G$ acts transitively on $A$, then there is only 1 orbit and so this implies that $\mathcal{O}(a) = G$. Hence, we have that $|G| = \frac{|G|}{|Stab_G(a)|}$. This implies that $|Stab_G(a)| = 1$ and so $Stab_G(a) = \{identity\}$. Thus, since the stabilizer of an arbitrary element only has the identity element, we can say that for any $a \in A$, we can find elements $g\in G$ such that $ga = a$ as desired.
Is this a right approach or where am I missing something?
We do need to assume $|A|>1$ otherwise this doesn't really make sense.
As noted in the comments, your mistake is equating $|\mathcal{O}(a)|$ and $|G|$, which is only true in one special case (the action of $G$ on itself by left or right multiplication). You can however prove this without (explicitly using) Burnside's Lemma. I realised after writing this that I do actually use the same special case of Burnside's Lemma as in the linked answer. I think you'll struggle to find a more elementary proof though.
Consider $\sum\limits_{a\in A}|{\rm Stab}_G(a)|$ - if every $g\in G$ fixes some point then for each $g\in G$ there is some $a\in A$ with $g\in{\rm Stab}_G(a)$. Moreover $1\in{\rm Stab}_G(a)$ for each $a\in A$ so $\sum\limits_{a\in A}|{\rm Stab}_G(a)|>|G|$. But $|{\rm Stab}_G(a)|=\frac{|G|}{|A|}$, so $\sum\limits_{a\in A}|{\rm Stab}_G(a)|=|G|$. A contradiction! Hence there must be some $g\in G$ which does not fix any point.
I don't cite Burnside's Lemma, but $\sum\limits_{a\in A}|{\rm Stab}_G(a)|=|G|$ is really a special case of Burnside's Lemma.