Q:

Probability of independent events, quick question

Quick question, if I have two independent events say $A$ and $B$ then I can write one event, such as $A$ as:
$A=A $\ $B$ , likewise $P(A) =P(A $\ $B$) 

Correct?

A:

(TLDR: No, not for independent events, but yes for mutually exclusive events)

To reiterate what is said in comments:
We notate set difference with $\setminus$.  That is to say $A\setminus B = \{x~:~x\in A,x\notin B\}$.  We would have then $Pr(A\setminus B)$ is the probability that $A$ occurs and $B$ does not occur simultaneously.
We notate conditional probability with $\mid$.  That is to say $Pr(A\mid B)$ denotes the probability that $A$ has occurred given knowledge that $B$ has occurred.
Given two events $A$ and $B$ of a probability space, the following statements are equivalent:

The two events $A$ and $B$ are said to be mutually exclusive (to one another).
$A\cap B = \emptyset$
$A\setminus B = A$
$B\setminus A = B$

In layman's terms, if $A$ and $B$ are mutually exclusive that means that they cannot happen at the same time.  In particular, if it is known that $B$ occurred then it is impossible for $A$ to have also occurred simultaneously, i.e. $P(A\mid B)=0$.

Given two events $A$ and $B$ of a probability space, the following statements are equivalent:

The two events $A$ and $B$ are said to be independent (of one another).
$Pr(A\cap B)=Pr(A)\times Pr(B)$
$Pr(A\mid B) = Pr(A)$
$Pr(B\mid A) = Pr(B)$

In layman's terms, independent events have no influence over one another.  Even if we know that $B$ has occurred, this gives us no knowledge and does not influence our guess as to whether or not $A$ has occurred (almost the exact opposite scenario as for mutually exclusive events).

You asked in your question whether or not given that $A$ and $B$ are independent if $A=A\setminus B$ and $Pr(A)=Pr(A\setminus B)$.  No, at least not so long as $B$ occurs with probability greater than zero.  These would have been true for mutually exclusive events but not for independent events.  I cannot stress enough that mutual exclusivity and independence are totally different and should never be confused with one another.
Instead, $Pr(A\setminus B) = Pr(A\cap B^c) = Pr(A)\times (1-Pr(B))\neq Pr(A)$ so long as $Pr(B)\neq 0$

