Group-Isomorphism problem I want to find an group-isomorphism $$ \psi : (\mathbb{Z}/8\mathbb{Z},+) \longrightarrow \mathbb{F}_9^\times $$
which should be used to multiply elements in $\mathbb{F}_9$ or to find the inverse element in an easy way.

As Seth and Pedro indicated the existence of such an isomorphism follows from (and is equivalent to) the cyclicity of the multiplicative group $\Bbb{F}_9^*$. To exhibit an explicit isomorphism you need to specify a construction of $\Bbb{F}_9$ and find a generator of the multiplicative group (aka a primitive element).
Elements of order $8$ are zeros of the cyclotomic polynomial $\phi_8(x)=x^4+1$. As we are working modulo $3$, we can factor it as follows
$$
\begin{aligned}
x^4+1&=x^4+4=(x^4+4x^2+4)-4x^2\\
&=(x^2+2)^2-(2x)^2=(x^2+2-2x)(x^2+2+2x)\\
&=(x^2-x-1)(x^2+x-1).
\end{aligned}
$$
So, if we identify $\Bbb{F}_9$ with $\Bbb{Z}_3[x]/\langle x^2-x-1\rangle$, then the coset
$\alpha=x+\langle x^2-x-1\rangle$ will be a generator.
The desired isomorphism $\psi:(\Bbb{Z}_8,+)\to (\Bbb{F}_9^*,\cdot)$ is given by
$$
\psi(\overline{n})=\alpha^n
$$
for all $n=0,1,\ldots,7$. This comes to the following:
$$
\begin{array}{c|ccc|c}
n&\psi(\overline{n})&&n&\psi(\overline{n})\\
\hline
0&1&&1&\alpha\\
2&\alpha^2=\alpha+1&&3&\alpha^3=\alpha^2+\alpha=2\alpha+1\\
4&\alpha^4=2&&5&\alpha^5=2\alpha\\
6&\alpha^6=2\alpha^2=2\alpha+2&&7&\alpha^7=\alpha+2
\end{array}
$$
Here I used the equation $\alpha^2=\alpha+1$ repeatedly. Note also that $\alpha^4=2=-1$, so
$\alpha^{4+j}=-\alpha^j$ for all $j$. As a final check we do that
$$
\alpha^8=\alpha\cdot\alpha^7=\alpha(\alpha+2)=\alpha^2+2\alpha=3\alpha+1=1
$$
as it should.
The way to use this in finding products and inverses is to treat the inverse of $\psi$ as a logarithm, i.e. a mapping that transforms multiplication in $\Bbb{F}_9$ into addition in $\Bbb{Z}_8$. For example (read the above table from right to left)
$$
(\alpha+2)(\alpha+1)=\alpha^7\cdot\alpha^2=\alpha^9=\alpha^{8+1}=\alpha,
$$
and
$$
(\alpha+1)^{-1}=(\alpha^2)^{-1}=\alpha^{-2}=\alpha^{8-2}=\alpha^6=2\alpha+2.
$$
You mentioned that you had found $x+2=x-1$ and $2x+1$ to be generators of $\Bbb{F}_9^*$.
That may be true, but to make that meaningful you need to specify the minimal polynomial of $x$. Above I used the minimal polynomial $x^2-x-1$, and $\alpha$ was one of its zeros ($\alpha^3=2\alpha+1$ being the other. If a cyclic group of order $8$ is generated by an element $c$, then $c^m$ is also a generator iff $\gcd(m,8)=1$. So here any of $\alpha$, $\alpha^3$, $\alpha^5$ and $\alpha^7$ is a generator. Not surprisingly those are the zeros of $x^4+1$ in this copy of $\Bbb{F}_9$.

Changing the irreducible polynomial always changes the scene. Your copy of $\Bbb{F}_9$ is $\Bbb{Z}_3[x]/\langle x^2+1\rangle$. Because $x^2+1\mid x^4-1$ in this case $x+\langle x^2+1\rangle$ will not work as a generator. If we let $\beta=x+\langle x^2+1\rangle$, then 
for $\alpha=\beta+2$ (that you might call $x+2$ even though that abuses notation slightly)
we get
$$
0+\langle x^2+1\rangle =(x^2+1)+\langle x^2+1\rangle=(\alpha+1)^2+1=\alpha^2+2\alpha+2=\alpha^2-\alpha-1.
$$
Thus $x^2-x-1$ is a minimal polynomial of $\alpha$. In terms of $\beta$ the above the homomorphism looks like
$$
\begin{array}{c|ccc|c}
n&\psi(\overline{n})&&n&\psi(\overline{n})\\
\hline
0&1&&1&\beta+1\\
2&(\beta+1)^2=2\beta&&3&(\beta+1)^3=2\beta^2+2\beta=2\beta+1\\
4&(\beta+1)^4=2&&5&(\beta+1)^5=2\beta+2\\
6&(\beta+1)^6=\beta&&7&(\beta+1)^7=\beta+2
\end{array}
$$
All the elements on the right column ($\beta+1,2\beta+1,2\beta+2,\beta+2$) can take the role of the generator. Thus you get four different such isomorphisms $\psi$. Each will work equally well as the inverse of a discrete logarithm.
