I am trying to prove that φ(a) = φ(b) if and only if a*Ker (φ) = b * Ker( φ).
This is what I have so far:
φ(b)^(-1) φ(b)^(-1) φ(a)(b)= φ(b)^(-1) φ(a)
e' = φ(b)^(-1) φ(a)
e' = φ(b^-1) φ(a) therefore (b^(-1) * a) is an element of the kernal
and b* kerφ = a* kerφ
and going in the other direction I have
a* kerφ = b* kerφ
ae' = be'
φ(a)=φ(b)
Will this proof work?
No, I don't think it will.
Let's start with part (2):
let's call $\text{ker }\phi, K$.
So, given $aK = bK$, we know that $a = ae = bk$, for some $k \in K$.
Thus $\phi(a) = \phi(bk) = \phi(b)\phi(k)$, since $\phi$ is a homomorphism.
Given that $\phi(k) = e'$ (since $k \in \text{ker }\phi$), what are your conclusions?
The first part is a bit better, but it should start like this:
$\phi(a) = \phi(b) \implies [\phi(b)]^{-1}\phi(a) = e'$
Your first statement is a bit muddled, and has too many instances of $\phi(b)^{-1}$.