I was not able to find a beginner introduction to superstructures and the cumulative hierarchy that makes me able to answer to some of my questions about them so I tried to ask here and I apologize if the notation is not the standard in this argument
Let's define $V_\omega$, $\mathcal V^{B}_\omega$, $\mathcal V^{*B}_\omega$ and $\mathcal S^{B}_\omega$
$$V_0:=\varnothing$$
$$V_{\alpha+1}:=\mathcal P(V_\alpha)$$
$$\bigcup_{\alpha \in \omega} V_\alpha:=V_\omega$$
Then I define $\mathcal V^{B}_\omega$ as a "weak superstructure" defined from a set of $B$ of base elements (urelements): I don't now the terminology but I think I should only use a set $B$ that doesn't already belong to $V_{\alpha<\omega}$ because in this case $\mathcal V_\omega ^B=V_\omega$
$$\mathcal V_0^B:=B$$
$$\mathcal V_{\alpha+1}^B:=\mathcal P(\mathcal V^B_\alpha)$$
$$\bigcup_{\alpha \in \omega} \mathcal V^B_\alpha:=\mathcal V^B_\omega$$
$\mathcal V^{*B}_\omega$ is defined like $\mathcal V^{B}_\omega$ but using the powerset without empty set $\mathcal P^*(X):=\mathcal P (X)\setminus \{\varnothing\}$ instead of the usal powerset.
$$\mathcal V_0^{*B}:=B$$
$$\mathcal V_{\alpha+1}^{*B}:=\mathcal P^*(\mathcal V^{*B}_\alpha)$$
$$\bigcup_{\alpha \in \omega} \mathcal V^{*B}_\alpha:=\mathcal V^{*B}_\omega$$
And at the end I define $\mathcal S^{B}_\omega$ as the superstructure of a set $B$ of urelements in this way
$$\mathcal S_0^B:=B$$
$$\mathcal S_{\alpha+1}^B:=\mathcal P(\mathcal S^B_\alpha) \cup \mathcal S^B_\alpha$$
$$\bigcup_{\alpha \in \omega} \mathcal S^B_\alpha:=\mathcal S^B_\omega$$
$1$-what are the relations betwen these three definitions?
I notice that $\mathcal V^\varnothing_\omega=\mathcal S^\varnothing_\omega=V_\omega$
And if $B$ is generated at some levels of $V_\omega$ then $\mathcal V^B_\omega \subseteq V_\omega$ but if I'm not wrong $V_\omega \subseteq \mathcal V^B_\omega$ too so in that case $\mathcal V^B_\omega = V_\omega$ is this right?
$2$-I defined $\mathcal V^{*B}_\omega$ in order to obtain a pure version of $\mathcal V^{B}_\omega$, with pure I mean without all the sets that have nothing to do with $B$ so I would like to know if, as I defined it, the following holds
$$\mathcal V^{*B}_\omega=\mathcal V^{B}_\omega\setminus V_\omega$$
and how is called this kind of "weak superstructure".
The last is about the inclusion of some of these "superstructures" the following
$3$-Is inclusion is true
$$V_\omega \subseteq \mathcal V^{B}_\omega \subseteq \mathcal S^{B}_\omega $$
There are other fundamental properties that I didn't noticed that are usefull when working witht these constructions?
I'll leave hints for now in the spirit of pedagogy. Details will be filled in at a later time or upon request.
A useful property for confirming your hunches in 1 (and part of 3) is that $A \subseteq B \implies \mathcal V_\omega^A \subseteq \mathcal V_\omega^B$ (and likewise for $\mathcal V_\omega^*$ and $\mathcal S_\omega$). See if you can convince yourself that this is the case.
1: Explicitly, you're trying to prove that if $B \subseteq V_\alpha$ for some $\alpha \in \mathbb Z_{\geq 0}$, then $V_\omega \subseteq \mathcal V_\omega^B \subseteq V_\omega$. The property $A \subseteq B \implies \mathcal V_\omega^A \subseteq \mathcal V_\omega^B$ from above is useful here, as are some of the progress you've already made in your question!
3: Yes. The first relation was already proved as 1. The second can be proven by induction and properties of unions.
2: The claimed equation $\mathcal V_\omega^{*B} = \mathcal V_\omega^B \setminus V_\omega$ is false in general: The situation in part 1 already furnishes counterexamples. But you did write that you would prefer for $B$ to consist of urelements. In that case, consider letting $B$ be something simple like $\{x\}$ or $\{x,y\}$, where $x$ and $y$ are urelements. What do the elements of $\mathcal V_\omega^{*B}$ look like? What about $\mathcal V_\omega^B \setminus V_\omega$?
As for names for $\mathcal V_\omega^B$ and $\mathcal V_\omega^{*B}$, I can't seem to find any (other than something generic like "union of $n$-fold power sets of $B$"). Though perhaps you shouldn't consider that authoritative since I'm a beginner in this topic.