Determine invertible elements of a monoid I have the following excercise.

In the set $\mathbb{Z}_6\times\mathbb{Z}_6$ consider the follow:
  $$(\overline{a}, \overline{b})\cdot(\overline{c}, \overline{d}) = (\overline{a}+\overline{c}+\overline{3}, \overline{b}\overline{d})$$
   Prove that $(\mathbb{Z}_6\times\mathbb{Z}_6, \cdot)$ is a monoid,
  and determine the invertible elements.

Yes, it's a monoid.
Associative:
$$\big((\overline{a}, \overline{b})\cdot(\overline{c}, \overline{d})\big) \cdot  (\overline{e}, \overline{f})=(\overline{a}, \overline{b})\cdot\big((\overline{c}, \overline{d}) \cdot  (\overline{e}, \overline{f})\big)$$
through various steps (and considering $\overline{3}+\overline{3}=\overline{6}=\overline{0} $, is that correct???):
$$(\overline{a}+\overline{c}+\overline{e}, \overline{b}\overline{d}\overline{f})=(\overline{a}+\overline{c}+\overline{e}, \overline{b}\overline{d}\overline{f})$$
Identity:
 $$(\overline{a}, \overline{b})\cdot(\overline{e_1}, \overline{e_2}) = (\overline{a}, \overline{b})\quad \forall (\overline{a}, \overline{b}) \in \mathbb{Z}_6\times\mathbb{Z}_6$$
follow:
$$(\overline{a}+\overline{e_1}+\overline{3}, \overline{b}\overline{e_2})=(\overline{a}, \overline{b})$$
thus:
$$\overline{a}+\overline{e_1}+\overline{3} = \overline{a}\Rightarrow \overline{e_1}=\overline{-3}$$
and
$$\overline{b}\overline{e_2}= \overline{b}\Rightarrow \overline{e_2}=\overline{1}$$
The structure is even commutative, than exist right and left identity.
Invertible elements:
$$(\overline{a}, \overline{b})\cdot(\overline{i_1}, \overline{i_2}) = (\overline{-3}, \overline{1})\quad \forall (\overline{a}, \overline{b}) \in \mathbb{Z}_6\times\mathbb{Z}_6$$
so:
$$(\overline{a}+\overline{i_1}+\overline{3}, \overline{b}\overline{i_2})=(\overline{-3}, \overline{1})$$
follow (and considering: $\overline{3}+\overline{3}=\overline{6}=\overline{0}$):
$$\overline{a}+\overline{i_1}+\overline{3} = \overline{-3}\Rightarrow \overline{a}+\overline{i_1}+\overline{3} +\overline{3} = 0 \Rightarrow \overline{i_1}=\overline{-a}$$
and
$$\overline{b}\overline{i_2}=\overline{1} \Rightarrow \overline{i_2}=\overline{b^{-1}}$$
Is (until now) all right? Now, how can I find $\overline{b^{-1}}$?

Since you are in $\mathbb Z_6, \overline{b}^{-1}$ only exists for $\overline b= \overline 1$ or $\overline b= \overline5$.
