I am following the correct proof is given by Marton. But why is the following incorrect ?
F and (E-F) are mutually disjoint and union is $E\cup F$
Therefore Eq1 is $(E\cup F)-(E-F)=F$
(F - E ) and $(E\cap F)$ are independent and Eq2 is
$F- (E\cap F) = F - E$
Eq1 - Eq2 is
$(E\cup F) + (E\cap F) = 2E$
Eq$1$ and Eq$2$ are true but you cannot add or substract "set equations". Kind of why I prefer $E \setminus F$ to $E - F$. Instead you need to think about why set equalities are true, for example by checking both inclusions.
In general, set "addition" is not defined, since the more natural "addition" on sets is the union.
When the sets are taken over spaces on which addition or multiplication or [insert operation] (e.g. groups, rings, vector spaces, etc...) you can start to define sets like $A + B := \{a + b \mid a \in A, b \in B\}$, $\lambda A := \{\lambda a \mid a \in A\}$, $AB := \{ab \mid a \in A, b \in B\}$ and so forth (for convenience) but they will not obey any particular algebraic operation in the way that you'd expect your numbers, elements, vectors, etc... to obey: you can have $A + A = A$, $-A = A = 3A$, $AB = A$, $A + (- A) \neq \emptyset$, and so on.
And that's fine, because that is not what is asked of sets. But that does mean one needs to be more careful when dealing with such notations.
EDIT: Once you go and use a measure like in the link you gave, your equations will make more sense:
Let $\mathbb{P}$ be a probability measure on a measurable space $(\Omega, \mathcal{A})$, and let $(E,F) \in \mathcal{A}^2$.
Then we have, thanks to $\mathbb{P}(A \setminus B) = \mathbb{P}(A) - \mathbb{P}(A \cap B)$ : $$\begin{split}\mathbb{P}(F) &= \mathbb{P}((E \cup F) \setminus (E \setminus F))\\ &= \mathbb{P}(E \cup F) - \mathbb{P}(E \setminus F)\\ &= \mathbb{P}(E \cup F) - (\mathbb{P}(E) - \mathbb{P}(E \cap F))\end{split}$$ And thus: $$\mathbb{P}(E \cup F) = \mathbb{P}(E) + \mathbb{P}(F) - \mathbb{P}(E \cap F)$$ So you can prove what Marton showed with set substractions rather than disjoint unions if you want to. However the advantage of Marton's approach is that their method does generalise to measures that can take $+\infty$ as a possible value like the counting measure I mentioned in the comments below, because it only deals with adding positive possibly infinite numbers, while this approach requires the measure to only take finite values.
Also I slightly misjudged your equations, you would have needed to turn one of $E \setminus F$ or $F \setminus E$ to the other one, so that they simplify each other once substracting the equations (once the equations are rewritten with measures that is, not right now).