Find the identity under a given binary operation I have two problems quite similar. The first:

In $\mathbb{Z}_8$ find the identity of the following commutative operation:
  $$\overline{a}\cdot\overline{c}=\overline{a}+\overline{c}+2\overline{a}\overline{c}$$

I say:
$$\overline{a}\cdot\overline{i}=\overline{a}+\overline{i}+2\overline{a}\overline{i} = \overline{a}$$
Since $\overline{a}$ is always cancelable in $\mathbb{Z}_9$ I can write:
$$\overline{i}+2\overline{a}\overline{i} = \overline{0}$$
$$\overline{i}(\overline{1}+2\overline{a}) = \overline{0}$$
so $i=0$ whatever $\overline{a}$.
Second question:

In $\mathbb{Z}_9\times\mathbb{Z}_9$ find the identity of the following
  commutative operation: $$(\overline{a}, \overline{b})\cdot
 (\overline{c}, \overline{d})= (\overline{a}+ \overline{c},
 \overline{8}\overline{b}\overline{d})$$

So starting from:
$$(\overline{a}, \overline{b})\cdot
 (\overline{e_1}, \overline{e_2})= (\overline{a}+ \overline{e_1},
 \overline{8}\overline{b}\overline{e_2})=(\overline{a}, \overline{b})$$
that is:
$$\overline{a}+\overline{e_1}=\overline{a}\qquad (1)$$
$$\overline{8}\overline{b}\overline{e_2}=\overline{b}\qquad (2)$$
In (1) there's always cancellable element for $\overline{a}$ since $\overline{-a}$ is always present in $\mathbb{Z}_9$.
In (2) I should multiply both member for $\overline{8^{-1}}$ and $\overline{b^{-1}}$ to  know exactly $\overline{e_2}$.
This happens only if both number are invertible. $\overline{8}$ is easily to demonstrate that it's invertible, cause $gcd(8,9)=1$.
But what about $b$.

Hint $\ $ Identity elements $\rm\:e\:$ are idempotent $\rm\:e^2 = e\:$. Therefore
$\rm(1)\ \ mod\ 8\!:\ \ e = e\cdot e = 2e+2e^2\ \Rightarrow\ e\:(1+2e) = 0\ \Rightarrow\ e = \ldots\:$  by $\rm n^2 \equiv 1\:$ for $\rm\:n\:$ odd
$\rm(2)\ \ mod\ (9,9)\!:\ \ (a,b) = (a,b)^2 = (2a,-bb)\ \Rightarrow\ (-a, b\:(b+1)) = (0,0)\ \Rightarrow\ \ldots$
