Prove that $G$ be a group under $\oplus$ that defined by $\bar{a} \oplus \bar{b} = \bar{a} \times_7 \bar{b} +_7 \bar{a} +_7 \bar{b}$. In this problem, I stuck when find the inverse of each element in $G$. Please help me at least give a hint so that I can solve this problem clearly.
Problem

Let $\mathbb{Z}_7$ be a group under $+_7$ and $\mathbb{Z}_{7}^{*}$ be a group under $\times_7$ where $\mathbb{Z}_{7}^{*} = \lbrace \bar{a} \in \mathbb{Z}_7 \mid \bar{a} \neq \bar{0} \rbrace$. Let a nonempty set $G$ that defined as $G = \lbrace \bar{a} \in \mathbb{Z}_7 \mid \bar{a} \neq \bar{6} \rbrace$ and a binary operation $\oplus$ on $G$ that defined by $$\bar{a} \oplus \bar{b} = \bar{a} \times_7 \bar{b} +_7 \bar{a} +_7 \bar{b}$$ for all $\bar{a},\bar{b} \in \mathbb{Z}_7$. Prove that $G$ is a group under $\oplus$.

My Solution.
It's easy to show that $\oplus$ is an associative binary operation.
Now, $\bar{0} \in G$. Then, $\bar{0} \oplus \bar{a} = \bar{0} \times_7 \bar{a} +_7 \bar{0} +_7 \bar{a} = \overline{0+a} = \bar{a} = \bar{a} \oplus\bar{0}$. Thus, $\bar{0}$ be an identity element of $G$.
Next, we'll find the inverse. Let $\bar{a}, \bar{m} \in G$ where $\bar{m}$ be the inverse of $\bar{a}$. Then, $\bar{0} = \bar{m} \oplus \bar{a} \Rightarrow \bar{m} = -\frac{\bar{a}}{\bar{a}+1} \notin G$.
I get stuck. Please help at least give me some hint. Thanks!
 A: Associativity follows from that of multiplication & addition of integers.
It is clear that, indeed, $\bar{0}$ is the identity.
The Cayley table is then computed, with the help of $\color{blue}{\text{commutativity}}$ of multiplication & addition of integers (and hence of $\oplus$), as follows:
$$\begin{array}{c|cccccc}
\oplus & \bar{0} & \bar{1} & \bar{2} & \bar{3} & \bar{4} & \bar{5} \\
\hline
\bar{0} & \bar{0} & \bar{1} & \bar{2} & \bar{3} & \bar{4} & \bar{5} \\
\bar{1} & \color{blue}{\bar{1}} & \bar{3} & \bar{5} & \bar{0} & \bar{2} & \bar{4} \\
\bar{2} & \color{blue}{\bar{2}} & \color{blue}{\bar{5}} & \bar{1} & \bar{4} & \bar{0} & \bar{3} \\
\bar{3} & \color{blue}{\bar{3}} & \color{blue}{\bar{0}} & \color{blue}{\bar{4}} & \bar{1} & \bar{5} & \bar{2} \\
\bar{4} & \color{blue}{\bar{4}} & \color{blue}{\bar{2}} & \color{blue}{\bar{0}} & \color{blue}{\bar{5}} & \bar{3} & \bar{1} \\
\bar{5} & \color{blue}{\bar{5}} & \color{blue}{\bar{4}} & \color{blue}{\bar{3}} & \color{blue}{\bar{2}} & \color{blue}{\bar{1}} & \bar{0} 
\end{array},$$
from which one can deduce that 
$$\begin{align}
\bar{1}^{-1}&=\bar{3},\\
\bar{2}^{-1}&=\bar{4},\\
\bar{3}^{-1}&=\bar{1},\\
\bar{4}^{-1}&=\bar{2},\,\text{ and}\\
\bar{5}^{-1}&=\bar{5}.
\end{align}$$
Closure is also implied by the table.
Thus $(G, \oplus)$ is a group.
