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 $b$.

So since you want $$(\overline{a},\overline{b})=(\overline{a}+\overline{e_1},\overline{8}\overline{b}\overline{e_2})$$
You must have $\overline{a}=\overline{a}+\overline{e_1}$, meaning $\overline{e_1}=\overline{0}$, and from the second entry we see that 
$\overline{b}=\overline{b}(\overline{8}\overline{e_2})$; since this holds for any $\overline{b}$ we must have $\overline{8}\overline{e_2}=1$, and as you said $\overline{8}$ has a multiplicative inverse, so you have that $\overline{e_2}=\overline{8}^{-1}(1)=\overline{8}^{-1}=\overline{8}$ since $8\cdot 8=64=1$ modulo $9$. So your answer is $\overline{e_1}=\overline{0}$ and $\overline{e_2}=\overline{8}$. 
