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Def. If $A$ and $B$ are sets of sets, let $A\diamond B = \{X \bigtriangleup Y\mid X\in A,Y\in B\}$, where $X\bigtriangleup Y$ is the regular symmetric difference of two sets.

Def. A finite set of sets $A$ is saturated (ad hoc name) if it has a maximum, a minimum, and the additional property that if $X,Z\in A$, then any $Y$ such that $X\subseteq Y\subseteq Z$ is also in $A$.

Example. If we write $ab$ instead of $\{a,b\}$, and $A=\{ab, c\}$, $B=\{a, b\}$, then $A\diamond B = \{b, ac, a, bc\}$. The set of sets $\{\emptyset, a, bc, abc\}$ is not saturated because $a\subseteq ab\subseteq abc$, but $ab$ is not in it. However, $\{\emptyset, a, b, c, ab, ac, bc, abc\}$ is. Yes, the lattice of subsets of some given set is a good model for a saturated set.

The following claim, now, can be shown to be true.

Claim. If $A$ and $B$ are saturated, then $A\diamond B$ is also saturated.

Example. If $A=\{\emptyset, a, b, ab\}$ and $B=\{b, bc, bd, bcd\}$, then $A\diamond B$ looks as follows:

\begin{array}{c|cccc} \diamond & b & bc & bd & bcd\\\hline \emptyset & b & bc & bd & bcd\\ a & ab& abc& abd& abcd\\ b & \emptyset& c& d & cd \\ ab & a & ac & ad & acd \\ \end{array}

It can be shown straightforwardly that $A\diamond B$ is closed under intersection, is closed under union and contains every set in between some $X, Y\in A\diamond B$. For instance, if we take some $X\bigtriangleup X', Y\bigtriangleup Y' \in A\diamond B$, then we can actually construct some $Z\in A$ and $Z'\in B$, based on $X,X',Y,Y'$, such that $Z\bigtriangleup Z' = (X\bigtriangleup X')\cap (Y\bigtriangleup Y')$. And so on, for the other operations. But this is tedious and does not seem to me very informative.

So my questions are as follows.

Q1. What kind of algebraic structure characterizes saturated sets as defined above?

I suppose there is some established name for it, but I couldn't quite find it.

Q2. Is there some algebraic argument why the claim presented above holds?

I suppose, also, that there is some deeper reason why the claim is true, and I'd like to understand it.

Q3. Out of curiosity, do things stay the same if $A$ and $B$ are infinite?

I don't see why not.

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    The sets you call saturated (which seems like a fine name for them) are precisely those of them form $\{X\subseteq A\mid B\subseteq X\}$ for some set $A$ and subset $B\subseteq A$. – 2017-02-13
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    Equivalently, a set of the form $\{Y\cup B\mid Y\subseteq C\}$ for some disjoint sets $B$ and $C$. (Take $C=A\setminus B$.) – 2017-02-13
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    For an arbitrary partially ordered set, there is the notion of an [interval](https://en.wikipedia.org/wiki/Partially_ordered_set#Interval). Your "saturated" sets are just closed intervals in the subset ordering. – 2017-02-13
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    Ah, yes, thank you. What I was wondering is if such an interval is some kind of algebra, or if 'finite interval of subset ordering' is the most we can say abut it. – 2017-02-13

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As Tobias Kildetoft notes, a "saturated" family is simply one of the form $\{X\mid A\subseteq X\subseteq B\}$ for some $A\subseteq B$.

It seems to be slightly easier to represent saturated families uniquely as a pair of disjoint sets $A$ and $Q$, where the saturated family is then $$\mathsf S(A,Q) = \{X\mid A\subseteq X\subseteq A\cup Q\}$$ Then we can also write $$ \mathsf S(A,Q) = \{ A\cup D \mid D\in\mathcal P(Q) \}$$

In this representation it is fairly easy to see that $$ \mathsf S(A,Q)\diamond \mathsf S(A',Q') = \mathsf S((A\triangle A')\setminus (Q\cup Q'),\; Q\cup Q')$$ Verifying the inclusions both ways is routine, and ought to be possible at the set algebra level, which more or less takes care of your Q2.

Since we didn't assume that anything was finite, the answer to your Q3 is "yes".

As for your Q1, I'm not sure there is a structure to point to that will tell a lot about the behavior of your operation. It is clearly a commutative monoid with identity $\mathsf S(\varnothing,\varnothing)=\{\varnothing\}$, and if we restrict everything to happen within a universal set $U$, then $\mathsf S(\varnothing,U)=\mathcal P(U)$ is an absorbing element.

Beyond that, the algebraic properties of the $\triangle$ operation the $A$s combine with, and the $\cup$ operation the $Q$s combine with are so different that they probably don't combine into any nice properties of $\diamond$.