So Axiom of regularity states: every non-empty set A contains an element that is disjoint from A
I'm wondering if this is equivalent as any set is not a member of itself? If so, how do we prove it?
If not, I am wondering is knowing any set is not a member of itself sufficient to derive that a∈b and b∈a cannot be both true?
No, this is not equivalent.
It is consistent that $A\notin A$ for all $A$, but there is $A,B$ such that $A\in B\in A$. The proof is quite difficult, but we can start with a model with atoms, and an extensional relation on the atoms, then realize that relation as $\in$.
With this we can start with two atoms, and the relation will simply be $a\mathrel{E}b\mathrel{E}a$.
You can ask whether or not the axiom of regularity is equivalent to just saying that there are no infinite decreasing sequences, and without the axiom of choice this is not true either using the same constructions, only mixing in some counterexamples to the axiom of choice.