Nonsense from combining two iffs ($\iff$) E and F are independent iff $\frac{p(E \cap F)}{p(F)}=p(E)$. 
Also, E and F are independent iff $p(E | F)=p(E)$
Why do I get nonsense if I combine the two? I'd get: E and F are independent iff $p(E | F) = \frac{p(E \cap F)}{p(F)}$
Why doesn't the iff carry over when I combine the two equations?

[This answer assumes the OP's argument is different from that assumed by the other answers.]
There's a logical fallacy in your argument, which seems to be of the following form:


*

*$A\iff(x=y)$

*$A\iff(z=y)$, where we define $z::=x$

*Hence, $A\iff [(x=y)\land (z=y)]$  


So far, so good, but the next step, eliminating $y$ in the conjunction, is incorrect:


*Therefore, $A\iff(x=z)$.  


That step would be correct if 
$$[(x=y)\land (z=y)]\ \ \iff\ \  (x=z),
$$
but this fails simply because
$$[(x=y)\land (z=y)]\require{cancel}\ \ \cancel\impliedby\ \  (x=z).
$$ 
NB: In your case, $A$=("$E,F$ are independent"), $x=\frac{p(E \cap F)}{p(F)},$ $y=p(E),$ and $z=p(E\mid F).$
