Determine invertible and inverses in $(\mathbb Z_8, \ast)$ Let $\ast$ be defined in $\mathbb Z_8$ as follows:
$$\begin{aligned} a \ast b = a +b+2ab\end{aligned}$$
Determine all the invertible elements in $(\mathbb Z_8, \ast)$ and  determine, if possibile, the inverse of the class $[4]$ in $(\mathbb Z_8, \ast)$.
Identity element
We shall say that $(\mathbb Z_8, \ast)$ has an identity element if:
$$\begin{aligned} (\forall a \in \mathbb Z_8) \text { } (\exists \varepsilon \in \mathbb Z_8 : a \ast \varepsilon = \varepsilon \ast a = a)\end{aligned}$$
$$\begin{aligned} a+\varepsilon+2a\varepsilon = a \Rightarrow \varepsilon +2a\varepsilon = 0 \Rightarrow \varepsilon(1+2a) = 0 \Rightarrow \varepsilon = 0 \end{aligned}$$
As $\ast$ is commutative, similarly we can prove for $\varepsilon \ast a$.
$$\begin{aligned} a \ast 0 = a+0+2a0 = a \end{aligned}$$
$$\begin{aligned} 0\ast a = 0+a+20a = a\end{aligned}$$
Invertible elements and $[4]$ inverse
We shall state that in $(\mathbb Z_8, \ast)$ there is the inverse element relative to a fixed $a$ if and only if exists $\alpha \in (\mathbb Z_8, \ast)$ so that:
$$\begin{aligned} a\ast \alpha = \alpha \ast a = \varepsilon \end{aligned}$$ 
$$\begin{aligned} a+\alpha +2a\alpha = 0 \end{aligned}$$
$$\begin{aligned} \alpha(2a+1) \equiv_8 -a \Rightarrow \alpha \equiv_8 -\frac{a}{(2a+1)} \end{aligned}$$
In particular looking at $[4]$ class, it follows:
$$\begin{aligned} \alpha \equiv_8 -\frac{4}{(2\cdot 4+1)}=-\frac{4}{9} \end{aligned}$$
therefore:
$$\begin{aligned} 9x \equiv_8 -4 \Leftrightarrow 1x \equiv_8 4 \Rightarrow x=4 \end{aligned}$$
which seems to be the right value as
$$\begin{aligned} 4 \ast \alpha = 4 \ast 4 = 4+4+2\cdot 4\cdot 4 = 8 + 8\cdot 4 = 0+0\cdot 4 = 0 \end{aligned}$$
Does everything hold? Have I done anything wrong, anything I failed to prove?

From your post:

Determine all the invertible elements in $(\mathbb Z_8, \ast)$

You have shown that the inverse of $a$ is $\alpha \equiv_8 -\frac{a}{(2a+1)}$. I would go a step further and list out each element which has an inverse since you only have eight elements to check.
