Is the probability addition rule commutative? The probability of $A$ and $B$ is the intersection between the venn diagrams for $A$ and $B$. 
Then is $P(A \cap B) = P(B \cap A)$? 
If so, then surely $\frac{P(A)}{P(B)} = \frac{P(A|B)}{P(B|A)}$. 
Is that correct?
Let me rephrase the question.. Why is it always the case that the probability of A happening, then B happening is the same when the order is flipped?
 A: Keep in mind that $A\cap B = B\cap A$ per commutativity of set intersection. Then trivially $P$ should be equal for both as otherwise $P(X) \neq P(X)$.
A: You are certainly correct that $$\mathbb{P}(\text{A and B}) = \mathbb{P}(\text{B and A})$$
This just because the events "$A$ and $B$" and "$B$ and $A$" are exactly the same event, so by necessity they must have the same probability. 
(From now on I will denote $(B \text{ and } A)$ as $(B \land A)$, and analogously for $(A$ and $B) = (A \land B)$.)
The equation you derived from this also appears correct:
$$\frac{\mathbb{P}(A)}{\mathbb{P}(B)} = \frac{\mathbb{P}(A)\mathbb{P}(B \land A)}{\mathbb{P}(B)\mathbb{P}(B \land A)} = \frac{\mathbb{P}(A\land B)}{\mathbb{P}(B)} \cdot \frac{\mathbb{P}(A)}{\mathbb{P}(B \land A)} = \frac{\mathbb{P}(A|B)}{\mathbb{P}(B|A)}  $$
Moreover, the above equation only holds when $\mathbb{P}(B \land A)>0$. Otherwise $$\frac{\mathbb{P}(B \land A)}{\mathbb{P}(B \land A)}=\frac{0}{0}$$ which is undefined.
In any case, this does not contradict Bayes' Rule.
