Prove that $a-b=b-a\Rightarrow a=b$ without using properties of multiplication. Yesterday my Honors Calculus professor introduced four basic postulates regarding (real) numbers and the operation $+$:

(P1) $(a+b)+c=a+(b+c), \forall a,b,c.$
(P2) $\exists 0:a+0=0+a=a, \forall a.$
(P3) $\forall a,\exists (-a): a+(-a)=(-a)+a=0.$
(P4) $a+b=b+a, \forall a,b.$

And of course, we can write $a + (-b) = a-b$. Then he proposed a challenge, which was to prove that $$a-b=b-a\iff a=b$$ using only these four basic properties. The $(\Leftarrow )$ is extremely easy and we can prove using only (P3), but I'm struggling to prove $(\Rightarrow )$ and I'm starting to think that it is not possible at all.
My question is how to prove $(\Rightarrow )$, or how to prove that proving $(\Rightarrow )$ isn't possible, using only (P1), (P2), (P3), (P4)?

You are right, you can't only from those axioms, and here is why.
Consider $A=\mathbb{Z}/2\mathbb{Z}$. Then $1-0=1=0-1$. But $0 \neq 1$.
But you can show this is true if $2 \neq 0$ and your ring is a integral domain, where $2:=1+1$. In fact, 
$a-b=b-a \implies a=b-a+b \implies 0=b-a+b-a \implies 0=2b-2a \implies 0=2(b-a) \implies b-a=0 \implies b=a. $
