Proof of $x*0=0$ is invalid? My professor told me today that while my logic was good and all my steps after the assumption were correct, my argument was invalid because I was assuming what I want to prove.  I don't see how.  Neither do any of my peers.  If someone could explain it to me, I would be greatly appreciative.
Using the axioms of ring theory, prove $b*0 = 0$.
First, I proved that 0*1 = 0.  Then I did the following.  
$Proof$: Suppose $a*b=a $ for all $b \in \mathbb{R}$.   (This is where he said I made the error)
By Additive Identity we have $(a+0)*b = a$.
By Distribitive Property we have $a*b + 0*b = a$.
By assumption, $a*b = a$, so we have $a+ 0*b = a$.
So $(-a)+a+0*b = (-a)+a$.
By Associativity, we have $(-a+a)+0*b = (-a+a)$.
Then, $0+0*b = 0$ by Additive Inverses.
So $0*b = 0$ by Additive Identity.
By Commutativity of Multiplication, $0*b = b*0$, so $b*0 = 0$.
As I have been taught, we are allowed to make a basic assumption when writing proofs and then see what follows.  What I have done is prove the following statement.   
If $a*b =a$,  $\forall b \in \mathbb{R} $, then $a=0$.
I don't see how this is different than the hint we were given: If $a+b=a,$ then $b=0.$  In both cases, we are allowed to assume the if statement is true and then show that the result follows.  After talking to a lot of my peers, we are all of the same mind.   I think we would all benefit from a discussion as to why we cannot do this.
 A: You didn't show the asked property, all you've said is if there is an element "a" with such property then this element must be zero.
Here is a simple proof of it in terms of basic ring properties:
$$b\cdot 0 = b(0 + 0)= b\cdot 0 + b\cdot 0$$ 
$$\implies -(b\cdot 0) + b\cdot 0 = (-(b\cdot 0) + b\cdot 0) + b\cdot 0$$ 
$$\implies 0 = 0 + b\cdot 0= b\cdot 0$$ 
$$\implies 0=b \cdot 0$$
A: $b*0=b*(1-1)=b*1-b*1=b-b=0$
Long: Let $b \in \mathbb R$. Consider $b*0$. By additive identity, and inverses, we have that $0=1 + (-1)=1-1$. By distribution, we have that $b*0=b*1-b*1$ This of course simplifies to $b-b$ and by the additive inverse property, $b-b=0$
forgive me if I'm assuming anything that your prof wouldn't allow, but I'm pretty sure I appealed only to axioms and definitions. 
A: You got already two good answers, so I will concentrate on your statement:

As I have been taught, we are allowed to make a basic assumption when
  writing proofs and then see what follows.

A split second after I read this, I burst into laughter (pardon me), because I imagined the following surreal scene:

The prof. says: Ok you have to prove that the axioms imply the
  conclusion ($A \implies C$), but since you are all new to this game of
  proving, I will allow you to add one extra statement of your choice, to
  help you find the proof. But just one, and make sure it is not
  too outrageous.

:-)
Now I am sure your prof. did not say or mean that. So perhaps there was a miscommunication somewhere. Anyway it appears that you were not given some rudiments of proof theory, so let me show you what you did and why it was not correct.
Let's say that you are given the task to prove $A \implies C$ (the axioms imply the conclusion) and in the course of the proof you add - ie you assume - some auxiliary statements (Aux). Then you impeccably proceed with your proof and show that $C$ is true. Now what have you done? Have you really proved that $A \implies C$? No! What you have proved is simply $(A \& Aux) \implies C$. This reads: if A is true and Aux is true, then C is true. This is exactly what you did. 
Another, logically equivalent, way to describe what you did is that you proved $A \implies (Aux \implies C)$ which reads: if A is true then, if Aux is also true then C is true. 
As you see this theorem is much weaker than the one you were given to prove originally.
Now the question is: can we salvage (that is reinforce) a proof that assumes Aux, so that we can correctly say that we proved $A \implies C$? Yes, if we can prove that $A \implies Aux$. In this way we have shown that Aux is not an extravagant claim, but that it is simply a consequence of the axioms A.  In your case Aux is $a*b=a$ is very close to C ($b*0=0$) so proving $A \implies Aux$ is probably as tough as proving $A \implies C$ and not worth the effort. So in your case this strategy is not useful.
