Prove that $(ab = -b )\land (a\ne 0)\implies a+1=0$ I am having some troubles proving the implication that if $ab=-b$, and $b$ is not equal to $0$, then $a=-1$. I am only supposed to use axioms of integers to prove the proposition. I have written a proof, though I am not sure if it achieves its purpose. I am looking for some help to improve my proof where it may be weak.
Let $a,b$ be elements of integers. If $ab = -b$, then $a+1 = 0$
$ab = -b \implies ab+b = -b+b \implies ab+b = 0$
If $ab+b = 0$, then $ab+b = 0 = a+1$.
$a+1 = 0 \implies b(a+1) = 0 \implies ba+b = 0 \implies ab+b = 0$.
$ab+b = a+1 \implies b(a+1) = a+1 $
Here is where I get stuck. I need to prove $ab+b = a+1 \implies b(a+1) = a+1 \implies b=1$ Is there a way I can do so without division? So far I have this: $\implies (a+b)|(ba+b)=(a+b)|(a+b) \implies b=1$ Though I would like to avoid division if possible?
If $b=1$, then $ab=-b \implies a(1)=-(1) \implies a=-1$
As proven, $ab = -b$, then $a+1=0$.
If you have another approach to proving the implication, I'd really appreciate it if you could share it with me. I'm looking for some alternate methods to prove it.

$$ ab = -b $$
$$ ab + b = 0$$
Using converse of distributive property,
$$ b (a+1) = 0 $$
Now for $b(a+1)$ to be zero either $b$ or $a+1$ has to be zero. Since it is given that b is not zero so $a+1$ will be zero
$$ a+1=0\implies a = -1 $$
Hence proved that $a+1$ is zero as well as $a = -1$
