This is the exercise:
Let $A = \{a,b\}$ and list the four elements of the power set $\mathcal P(A)$. We consider the operations $+$ to be $\cup$, $\cdot$ to be $\cap$, and complement to be set complement. Consider $1$ to be $A$ and $0$ to be $\varnothing$.
Explain why the description above defines a Boolean algebra.
The elements of the $\mathcal P(A) = \{\{a\},\{b\},\{a,b\},\varnothing\}$.
Now I understand that in order to show that it defines a Boolean algebra, I need to show that the description meets the Boolean Algebra axioms of identity. What I don't understand, is how to show this. For example, one of the axioms is the identity axiom, which states that
$$x \cdot 1 = x$$
However, does this translate to $A \cap x = x$? and how is this true? As such, I was confused and could use some help.
The Boolean algebra in this case is the power set of $A$: $\mathcal P(A)$. Thus, in any axiom for Boolean algebras, such as $x\cdot 1 = x$, the $x$ is to be taken as an element of $\mathcal P(A)$, i.e. a subset of $A$ (not an element of $A$).
As mentioned in the comments, another useful thing to consider is that complement in this case is defined relative to $A$, so that $x^c = A \setminus x$.