Alternate Proof for the Cancellation Laws I will first state the theorem: 
Given a group $(G,*)$ the following laws apply for $a, b, c \in S $


*

*If $a*b=a*c$ then $b=c$

*If $b*a=c*a$ then $b=c$


Attempt at alternate Proof: 
Consider the contrapositive of (1), this would mean :

If $b$ is not equal $c$ then $a*b$ is not equal $a*c$

Proof: If $b$ is not equal $c$ ,  multiply $a$ on the right side of the inequality. This will imply that $a*b$ is not equal $a*c$.  
Is this correct? 
 A: In general, you cannot apply operations to both sides of an inequality.  For instance, $2\neq -2$ is a true inequality.  But if we square both sides of the inequality, we get $4\neq 4$, which is not true!
So in your situation, you cannot just take the inequality $b\neq c$ and multiply both sides by $a$ to get $a*b\neq a*c$.  You can't be sure that $b\neq c$ implies $a*b\neq a*c$ without giving some additional argument.  And if you try to give such an additional argument, I suspect you'll just end up coming up with a more direct proof of the original statement you were trying to prove.
Let me try to explain intuitively why you can apply an operation to both sides of an equation but not an inequality.  If you have an equation $x=y$, that means that $x$ and $y$ are literally the same thing; we're just referring to them by different names.  So any operation applied to $x$ will still be the same as any operation applied to $y$, because $x$ and $y$ are the same.
On the other hand, if $x$ and $y$ are different, there's no reason we should always expect them to remain different if we apply an operation to them.  Some operations can give the same output on different inputs.  Indeed, we saw an example of that with the squaring operation above.
A: The major point in showing that if $b\not=c$ then $ab \not= ac$ is to actually show that $ab\not=ac$, not just state it as in the question. The easiest is to prove the contrapositive by contradiction which is pretty much the same as any other standard proof of the original statement. So you can simply show that if $b\not=c$ and $ab=ac$ we obtain a contradiction. So, suppose that $b\not=c$. Then if $ab=ac$, we must have $b=eb=a^{-1}ab=a^{-1}ac=ec=c$, contrary to our assumption. 
If $G$ is finite, then I can think of another way:
Let $a \in G$. Define a map $\varphi_a:G \rightarrow G$ by $\varphi_a(x)=ax$. Then, given $y \in G$, we have $\varphi_a(a^{-1}y)=aa^{-1}y=y$ so that $\varphi_a$ is surjective. Since $G$ is a finite set, $\varphi_a$ is a bijection. Thus, if $b\not=c$, then $ab=\varphi_a(b) \not= \varphi_a(c)=ac$. 
The reason this doesn't work for infinite groups is that you can't prove that $\varphi_a$ is a bijection, unless you actually prove $ab=ac$ implies $b=c$.
This is simply the injectivity of the map $\varphi_a$.
A: You said: $$a\neq b\implies ca\neq cb$$
But consider this Monoid with operation $*$ :  
$$M=\{1,a,b\}\quad \text{s.t}\quad a\neq b\quad a*a=a*b=a\quad b*b=b*a=b$$
Now : 
$$a\neq b \implies ba=bb \quad,\quad aa=ab$$
What happened?!  

$M$ has all axioms of groups, except having the inverse element.

