How to prove $x-y = x+(-y)$ in ring theory. Okay, I have talked with a lot of people about this silly question. And I have thought about this way longer than is good for me. Everybody seem to disagree with me, and that is the reason I think I can't get this out of my head, because I feel like pure mathematically I'm right. 
If you look at the expression $x - y$ in ring theory. What does this mean? Okay of course, everybody here knows/agrees that this mean $x+(-y)$. Where $-y$ is the additive inverse of $-y$. And $+$ the binary operation "addition".
So what is your problem Kasper ? Well, if you just look at the definitions in ring theory, then I wouldn't be able to conclude this. There is no definition saying that $x-y=x+(-y)$ (in the book I read about ring theory). What I am able to conclude, is that if I define $z=-y$. Then:
$$x{-y}=xz=x⋅z=x⋅-y$$
Because it is defined (at least in my book) that $x\cdot y =xy$ where "$⋅$" is the binary operation "multiplication". As the symbol "$-$" is only defined in the context of $-x$ being the additive inverse of $x$, this is the only conclussion I can logically find. 
Talking with other mathematicians I get the impression I'm a little bit alone in this view, but I think I should be able to justify every step in a math proof, using definitions/theorems/axioms etc. And I shouldn't justify my step because I know already since highschool what is meant with $x-y$. Being able to trace back every math proof to axioms and definitions is part of the beauty of math in my opinion.
So, I would say you need to define "$-$" in the context of $-x$ (unary operation) where it denotes the additive inverse, and define "$-$" in the context of $x-y$ (binary operation) where it denotes $x+(-y)$. But I talk to my teacher, he says, no in ring theory you only need "$-$" as an unary operation. If I talk here in chatroom, they tell me the same.  

Okay, going even deeper in this problem then it is good for me. In a comment in other question:

Suppose you wrote the axioms for a group using "addition" as the
  operation suppressed instead of "multiplication". That is, every group
  has an "addition" operation. Or equivalently xy means (x+y). Then, the
  inverse axiom for groups says "for all x, there exists a -x such that
  x-x=0." Again, I've suppressed the addition operation, so x-x=0
  implicitly best gets read as meaning (x+-x)=0. So, no, you don't need
  to define a binary operation of subtraction for this sort of problem.
  You just need to recognize that x(y-z) has multiplication suppressed
  first and addition suppressed second.

This makes a little bit sense to me, but still. If you look at groups, and write $xy=x+y$, then you also write $x^{-1}$ instead of $-x$. And in ring theory they define $xy=x⋅y$ not $xy=x+y$ (at least in my book).
 A: Edited to match revised question.
When $x-y$ is defined, it is of course defined to be $x+(-y)$, but we often don’t bother to define it. After all, there is no actual need to define subtraction in a ring: it’s just a convenience, and you can do everything without it. Yes, a careful author of a textbook would explicitly define $x-y$ if he planned to use the notation, but failure to do so is a pretty minor failing: the definition is easy to pick up from context even if it’s not stated explicitly. I’d even go so far as to say that learning to pick things up from context is part of learning to read mathematics as it is actually written.
A: You need some definition of $x-y$ to prove anything about it.
It is most common, at the level where you can speak about abstract rings, to define $x-y$ to mean $x+(-y)$, as Brian writes.
But you can also use the grade-school definition: $x-y$ means the unique $a$ that solves $a+y=x$. We can then prove that $x+(-y)$ satisfies this definition:
$$ (x+(-y))+y = x+((-y)+y) = x+0 = x $$
and also that any solution to this equation must be unique, because if $a+y=x$ and $b+y=x$, then we have $a+y=b+y$, and then
$$ a = a+0 = a+(y+(-y)) = (a+y)+(-y) = (b+y)+(-y) = b+(y+(-y)) = b+0 = b $$
