Two natural extensions of every algebra. Extension to subsets or functions. I don't exactly know the technical meaning of extension, but I was thinking that given a set $A$ and an operation $*$ on it we can extend the set $A$ in a very natural way and thus extend any algerbaic structure on it:


*

*working on the subsets of $A$

$$A_P= (\mathcal P(A),*^p)$$
and for $S,T \in \mathcal P(A)$
$$S*^pT:=\bigcup_{s\in S,t\in T}\{s*t\}$$



*

*or on the functions in $A^A$. 



$$A_F= (A^A,*^f)$$
and for $f,g \in A^A$
$$f*^fg:=\{(x,f(x)*g(x)):\forall x \in A\}$$



$Q1$-In both cases we have that the new
  algerbaic structures have a "isomorphic copy" of the original
  structure inside them:  can we think that we have extended it "adding new
  elements"? Do we lose something of the original structure?

$sg(a):=\{a\}$ is an ijection from $A$ to $\mathcal P(A)$ and an isomorphism between $(A,*)$ and $(sg[A],*^p)$
in the same way $K(a):=\{(x,a):\forall x\in A\}\in A^A$ (the constant function) is an injection from $A$ to $A^A$ and an isomorphism between $(A,*)$ and $(K[A],*^f)$.
$K[A]$ is the image of the set $A$

$Q2$-Are these extensions the more "intuitive" possible for $(A,*)$? And what is the terminology for $A_P$ and $A_F$

The last problem is about the relationship of $A_P$ and $A_F$. I don't know if they are isomorphic-homomorphic (I guess that $A$ should be infinite if we want a bijection)?

$Q3$-What are the relations between these two extensions? Are these extensions usefull?

 A: If $R$ is a ring, then $R^R$ is also naturally a ring, with both addition and multiplication defined pointwise, and we may identify $R$ naturally with the subring of constant functions.  However, with respect to the question "Do we lose something of the original structure?", consider the following:


*

*Even if $R$ is an integral domain, $R^R$ will not be.  For example if $R = S \cup T$ is any partition of $R$ into disjoint subsets, let $f$ be a function that vanishes only on $S$, and $g$ a function that vanishes only on $T$.  Then $f \cdot g = 0$ despite the fact that neither $f$ nor $g$ is $0$.

*Similarly, even if $R$ is a field, $R^R$ will not be; the only functions on $R$ that have multiplicative inverses are functions that are always nonzero.


Regarding $A_P$, it strikes me that this definition is not too different from, for example, what we do when we define the sum of ideals in a ring.  But on the other hand we do not define the product of two ideals to simply be the set of products of pairs of elements from the two factors; rather, we have to then allow for all sums of such products as well.  So if this definition of $A_P$ is not really useful even for the very natural case of wanting to multiply two subsets of a ring, it is probably not very useful in more general cases.
A: Algebras with a single binary operation are usually referred to as magmas (in Bourbaki for example), and I'm going to follow this convention in my post. Most of my remarks apply more generally to any universal algebra though, with an arbitrary number of operations each taking an arbitrary number of arguments.
Firstly the powerset algebra you described isn't used explicitly a great deal to the best of my knowledge, though the product it uses it utilised a fair amount in studying monoids and groups. In general it seems to be more useful to restrict attention to certain substructures, for example ideals of rings (as mweiss mentioned) and submodules of modules, and to define the product as being the relevant substructure generated by the products of the elements.
As for properties: given a magma $M$, the powerset magma $\mathcal{P}(M)$ is associative iff $M$ is and commutative iff $M$ is. If $1$ is a multiplicative identity for $M$, then $\{1\}$ is a multiplicative identity for $\mathcal{P}(M)$. Inversion doesn't work quite so nicely: one obvious counterexample is that $a \cdot \emptyset \neq \{1\}$, no matter the value of $a$. Another problem is given a ring $R$, the product on $\mathcal{P}(R)$ isn't necessarily distributive over the sum, the best we can say is that $A(B+C) \subset AB + AC$.
Moving on to the algebra of functions, let $M$ be a magma with operation $\cdot$. First note that the magma $M^M$ doesn't really require the domain to be $M$. Your construction can be generalised by instead taking the domain to be some arbitrary set, $X$, giving you the magma $M^X$, with the product $\bullet$ defined by
\begin{equation}
(f\bullet g)(x) = f(x)\cdot g(x).
\end{equation}
This in turn is a special case of a product of magmas: say you have a family $(M_i)_{i \in I}$ of magmas with operations $\cdot_i$, then we can define the product magma $\prod_{i \in I} M_i$ to consist of the set of functions $f$ from the index set $I$ into $\bigcup_{i \in I} M_i$ such that for each $i$, $f(i) \in M_i$, along with the product $\bullet$ defined by
\begin{equation}
(f \bullet g)(i) = f(i) \cdot_i g(i).
\end{equation}
The function $f$ can be viewed as an $i$-tuple, $(f(i))_{i \in I}$, a list containing for each element of the index set $I$ an element of the corresponding magma. For example in the case that $I = \{1,2\}$, $\prod_{i \in I} M_i = M_1 \times M_2$ and the elements are equivalent to pairs of elements $(a_1,a_2)$, with $a_1 \in M_1$, $a_2 \in M_2$, and with the product $\bullet$ defined by
\begin{equation}
(a_1,a_2)\bullet (b_1,b_2) = (a_1 \cdot_1 b_1,a_2 \cdot_2 b_2).
\end{equation}
The product $\prod_{i \in I} M_i$ contains isomorphic copies of each of the $M_i$ (by fixing the remaining arguments). It also preserves their properties, if any algebraic identity (such as commutativity or associativity) holds for each of the $M_i$, then it holds for $\prod_{i \in I} M_i$. In particular, any algebraic identity that holds for a magma $M$ holds for the magma $M^X$. You have to be careful though: as mweiss mentioned the product of multiple fields (or domains) is not a field (or domain) in turn, the reason being that the multiplicative inverse identity $a \cdot a^{-1} = 1$ didn't hold in the first place for all elements (zero was exempt), and this creates even more unwanted exemptions in the product.
