I have to prove that if $(ab)^i = a^ib^i$ is valid for $3$ consecutive integers, then the group is abelian. 
I've found this answer in the internet, but I can't understand the part that says we can cancel $a$ from the left and $b$ from the rigth.
For example:
$(\color{red}{a^{-1}}a)b(ab)^i = (\color{red}{a^{-1}}a)b a^{i+1}b^{i+1} \implies b(ab)^i = a^{i}b^{i+1}$
But how can I cancel $b$? Also, could someone explain this proof for me?
I agree that the answer you found is phrased poorly. However, note that from $a^{i+1} b^{i+1} = (ab)(ab)^i = aba^i b^i$ we can get $$a^i b = b a^i \tag 1$$ by multiplication of $a^{-1}$ and $b^{-i}$ on the left and right, respectively. Doing the same algebra with the change in variable $i \to i+1$ we get $$a^{i+1} b = b a^{i+1} \tag 2 $$ Now take $ab$ and start by multiplying by a "clever" form of the identity element to see that $$ab = ab(a^ia^{-i}) \\ = a(ba^i)a^{-i} \\ = a(a^ib)a^{-i} \\ (a^{i+1}b)a^{-i} \\ = ba^{i+1}a^{-i} \\ = ba$$ where line three comes from a substitution using $(1)$ and line five comes from a substitution using $(2)$.
