What does it mean to prove that two systems have to same theorems? Lets say there is two systems $L_1$ and $L_2$.If I get asked:

Show that $L_1$ and $L_2$ have the same theorems. 

Is it sufficient to prove the axioms of $L_1$ using the axioms of $L_2$ or vice-versa ? Or do I have to show that a wf is a theorem of $L_1$ iff it is a theorem of $L_2$ ?
 A: It is indeed sufficient to show that each axiom of $L_1$ is derivable from the axioms of $L_2$ and vice versa. This implies that any wff is a theorem of $L_1$ iff it a theorem of $L_2$.
To see why assume $\phi$ is a theorem of $L_1$. The proof $\Pi$ of $\phi$ can only use finitely many axioms of $L_1$, say $\alpha_1, \ldots, \alpha_k$. If we have proofs $\Pi_1, \ldots, \Pi_k$  of $\alpha_1, \ldots, \alpha_k$ in $L_2$, we can combine $\Pi_1, \ldots, \Pi_k$ and $\Pi$ to give a proof of $\phi$ in $L_2$. So any theorem of $L_1$ is a theorem of $L_2$. Likewise, swapping $L_1$ and $L_2$ any theorem of $L_2$ is a theorem of $L_1$.
Aside: the above assumes that your systems have the same inference rules, e.g., if you are looking at a typical exposition of first-order logic, as a Hilbert style system with modus ponens as the only inference rule. If you are looking at systems with different inference rules (e.g, a natural deduction calculus on the one hand and a Hilbert-style system on the other), you will also have to show that each rule of $L_2$ is derivable (or more generally, what is called permissible) on the basis of the axioms and rules of $L_1$ and vice versa.
