Nonsense from combining two iffs ($\iff$)

You combined them wrong. From $A \Leftrightarrow B$ and $A \Leftrightarrow C$, you can infer $B \Leftrightarrow C$.

But you've misidentified what $A,B,C$ are. They aren't the sides of the equation, but rather

• $A$ is the statement "$E$ and $F$ are independent"
• $B$ is the formula $\frac{p(E \cap F)}{p(F)}=p(E)$
• $C$ is the formula $p(E | F)=p(E)$

so the correct combination is

$$ \frac{p(E \cap F)}{p(F)}=p(E) \qquad \mathrm{iff} \qquad p(E | F)=p(E)$$

Let $a=\frac{p(E \cap F)}{p(F)}$, $b=p(E | F)$, and $c=p(E)$. You know that the equations $a=c$ and $b=c$ are equivalent to each other (since they are both equivalent to "$E$ and $F$ are independent"). That is, whenever $a=c$ is true, $b=c$ is also true, and conversely.

There is no reason for $a=b$ to be equivalent to these two statements, though. If $a=c$ is true, then $b=c$ is also true, and so $a=b$ is true as well. So we have an implication in one direction. But the other direction doesn't work: if we know $a=b$, that doesn't tell us anything about whether $a=c$ or $b=c$ are true. It's entirely possible that $a$ and $b$ are equal to each other, but not to $c$.

Here is a simple mathematical example to clearly elucidate the logical error:

An integer $n$ is an even prime iff $n = 2$.

An integer $n$ is an even prime iff $n = 1+1$.

(WRONG!) An integer $n$ is an even prime iff $2 = 1+1$.