What makes it legitimate to multiply both sides? Having the proof of the cancelation law for multiplication:
$$cb=ab$$
$$(cb)b^{-1}=(ab)b^{-1}\tag{Inverse}$$
$$cbb^{-1}=abb^{-1}\tag{Associativity}$$
$$c\cdot 1=a\cdot 1\tag{Indentity}$$
$$c=a$$
This may seem stupid, but exactly at the second step: I can multiply both sides by the inverse, but is this a consequence of the field axioms or of the equality defined as an equivalence relation? I guess it's the second one, but I'm not sure if one could prove it with the field axioms alone. 
 A: Because if there were some operation you could perform on the two quantities to get different results, they would not be equal.
If two things are equal, the results of multiplying them each by $b^{-1}$, or of performing any other operation on them, must be equal, because that is what it means for two things to be equal.
A: One of my teachers I think seemed to dislike "multiplying" both sides of an equation by the same element. If you share the same aversion you can almost always use something like this to avoid it. 
$$c = c \cdot e = c \cdot(b \cdot b^{-1}) = (c \cdot b) \cdot b^{-1} = (a \cdot b) \cdot b^{-1} = a \cdot(b \cdot b^{-1}) = a \cdot e = a$$
All we have used here is the associative law for the product and substitution which is a much more fundamental concept than the definition of a group. 

But for the record I have no issues writing, 
$$ c = a \implies cb = ab $$
since all we use here is the fact that $$c= a \implies (c, b) = (a, b) \implies f(c, b) = f(a, b)$$ 
where $f : G \times G \to G$ is the binary product on $G$ which is a well-defined function. 
A: Yes, you can prove it from the properties of real numbers. 
Proof: 
If $a$, $b$, and $c$ are any real numbers, and $a = b$.
$ac$ is a unique real number $\iff$ closure property of multiplication 
$ac = ac$  $\iff$reflexive property of equality
$a = b$ given
$ac = bc$ $\iff$ substitution principle 
