As you've shown $Ker(\phi)=\{1,6\}$
You'd have $G/Ker(\phi)=\{\{1,6\},\{a,b\},\{c,d\}\}$
Note that $Ker(\phi)$ partitions the set $G$ in such a way that the corresponding pairs which are formed map to the same element under $\phi$.
Why?
By definition of cosets $G/Ker(\phi)=\{g.Ker(\phi)\,\,\big| \,\,g\in G\}$, then $\{a,b\}=\{g.1,g.6\}$ for some $g$.
Now $\phi(a)=\phi(g.1)=\phi(g)\phi(1)=\phi(g)\phi(6)=\phi(g.6)=\phi(b)$
Recall: $\phi(1)=1=\phi(6)$. Also note that $\phi(2)=4=\phi(5)$ and $\phi(3)=2=\phi(4)$ (NOTE that the operations still take place modulo $7$).
How did we reach here?
Well it is a well known fact that when we talk about ordinary addition $g^2=(-g)^2$
Now, same is the case here (with a little twist), what is the additive (modulo 7) inverse of $2$? $5$ is it? Of course, in fact additive inverse of $a$ (modulo 7) is $7-a$ and you have the same concept as the one used above.
$$a^2\equiv (7-a)^2\pmod7$$
So $\{a,7-a\}\in G/Ker(\phi)$ for $a\in H$ and the set becomes
$G/Ker(\phi)=\{\{1,6\},\{2,5\},\{3,4\}\}$
There is a more rigorous way that has been shown in most of the answers above, this is slightly different (intuitive?) way
Now, for the part where you ask what is $G/Ker(\phi)$ isomorphic to
Note $|G/Ker(\phi)|=|G|/|Ker(\phi)|=3$ and as it is a well known fact that any group (of quotients here) of prime order is cylic. So, $G/Ker(\phi)$ is isomorphic to any cyclic group of order 3. There is a theorem (Fundamental Theorem of Isomorphism) which states that
If $\phi:G\to G'$ (here G=G') is a homomorphism (Note that $G$ is onto $\phi(G)$), then $G/Ker(\phi)\cong\phi(G)$.
Now note, here $\phi(G)=\{1,2,4\}=\langle 4\rangle=\langle 2\rangle$ (modulo 7)
Remember that $e$ is an identity element in a group $(G,*)$ if for every $a$ in $G$ we have $a*e = a$.
For the first: $$*(a,e)=a\implies a+e-1=a \implies e=1$$
and second: $$*(a,e)=a\implies a+e+1=a \implies e=-1$$
so yes, your identity elements are correct.
Best Answer
Yes, the map $\phi$ is surjective (since it's image will be a non-trivial subgroup of $\mathbb{Z}_7$, and hence is all of $\mathbb{Z}_7$), and so the kernel is an index $7$ subgroup in $\mathbb{Z}$, but only $7\mathbb{Z}$ is such a subgroup.