Disclaimer: As pointed out in the comment section, I am giving an answer for $\alpha_2 \geq 0$. While writing the answer, I also noted that my solution requires $\Omega$ to be connected.
Step 1: $I(\cdot)$ is bounded below
For any $u \in X$, we have
$$
I(u)\geq \alpha_1||\nabla u||_{L^p}^p+\alpha_2||u||_{L^q}^q+\alpha_3\geq \alpha_3 >- \infty
$$
You already pointed out why $X$ is non-empty.
Step 2: Dealing with the minimizing sequence
So our functional is bounded below. Hence $m=\inf_{u \in X} I(u)$ exists. There exists a sequence $ \{ u_n \}_{n \in \mathbb{N}} \subset X$ such that $I(u_n)\to m$.
We can assume that the sequnce statsifies $M \geq I(u_n)$ for some $M>0$. Then, we can estimate
$$
M \geq I(u_n)\geq\alpha_1||\nabla u_n||_{L^p}^p+\alpha_2||u_n||_{L^q}^q+\alpha_3\geq \alpha_1||\nabla u_n||+\alpha_3
$$
This is possible, since $\alpha_2||u_n||_{L^q}^q\geq 0$ holds now.
Therefore, since $\alpha_1 >0$:
$$
||\nabla u_n||_{L^p}^p \leq \frac{M-\alpha_3}{\alpha_1}
$$
This gives us an upper bound on $||\nabla u_n||_{L^p}$.
Since all of our functions have a fixed mean value (as a property of $X$), you can now use a Poincare inequality involving mean values:
$$
||u_n||_{L^p}=||u_n-\alpha +\alpha||_{L^p}\leq ||u_n-\alpha||_{L^p}+||\alpha||_{L^p}\leq
C_1||\nabla u_n||_{L^p}+C_2
$$
A reference for this can be found, for example, in Evans PDE book. While writing the answer, I noted that we also need $\Omega$ to be connected.
Combining this with the bound on the gradient, you get that
$$
||u_n||_{W^{1,p}}\leq C
$$
Now we have a bounded sequnce in $W^{1,p}$, a reflexive Banach space. Now we can find a weakly convergent subsequence $u_k \rightharpoonup w$, where we have $w \in W^{1,p}$. It remains to show that $w \in X$. But this is due to Mazurs theorem. The set $X$ is convex and closed (why?) with respect to $||\cdot ||_{W^{1,p}}$ and hence weakly closed.
Step 3: Passing to the limit
You have already noted that this "uniform" convexity in $\zeta$ and boundedness below implies weak lower semicontinuity. This implies:
$$
I(w)\leq \liminf_{k \to \infty}I(u_k)=m=\inf_{u \in X} I(u)
$$
This leads to the conclusion:
$$
I(w)=\min_{u \in X} I(u)
$$
A few notes:
1) You did not need the Sobolev embedding at any point.
2) I left in a few gaps, since it would be way too much to write. A lot of theorems/steps use for example the properties of $\Omega$
3) The poincare inequality you were looking for was
$$
||u_n-\frac{1}{|\Omega|}\int_{\Omega}u_n(x)dx||_{L^p(\Omega)}\leq C||\nabla u_n||_{L^p(\Omega)}
$$
However, this needs the region $\Omega$ to be connected. I did forget about this while thinking about your post.
I hope this answer is still helpful and gives you an idea on how to deal with those problems.
4) I did not need the term $\alpha_2||u||_{L^q}^q$ at all. In fact, as $||u||_{L^q} \leq C_3 ||u||_{L^p}\leq C_4(||\nabla u_n||_{L^p}+1)$ makes the term look like it does not contain any valueable information. Now you might see what I tried: If $\alpha_2 < 0$, you could try to justify something like:
$$
\alpha_2||u||_{L^q} \geq \alpha_2 C_4(||\nabla u_n||_{L^p}+1)
$$
After some calculations, it will boil down to the sign of
$$
\alpha_1+C_5 \alpha_2
$$
This is basically impossible to predict, as $C_5=C_5(\Omega,p,n,q)\geq 0$.
Other techniques I know of were not applicable.
As a last and additional note for the $\alpha_2 < 0$ condition: You would need that for every sequence such that $\alpha_1 ||\nabla u_n||_{L^p}^p+\alpha_2||u_n||_{L^q}^q \to - \infty$ the functional stays bounded. Also, it all needs to be a sequence $X$, which means it has a fixed mean value.
