Is it the case that a proposition is equal to the product of all propositions implied by it? Is it the case that a proposition is equal to the product of all propositions implied by it?
More formally, it the case that
$$ \forall A. \left( A \leftrightarrow \underset{A \rightarrow B}{\bigwedge} B \right)$$
Also I think the logical dual should be something like
$$ \forall A. \left( A \leftrightarrow \underset{B \rightarrow A}{\bigvee} B \right)$$
but I can't quite make sense of that.
This seems right to me under some logics but not right for others. In particular for some incomplete logics it seems to me that there are some statements that are true if some statement is true but are not necessarily provably so. Basically, some statements $A \rightarrow B$ are not necessarily true or false but are unprovable either way and so it is not possible to get a sensible definition of the set of all such statements.

Bear in mind that the infinite conjunction:
$$\bigwedge_{A \to B}B$$
is not valid object language syntax in ordinary (finitary) first-order logic. If you treat it as a metanotation, you can explain it semantically as in the other answers. The natural way to view it syntactically is in second-order (propositional) logic, where it would be written thus:
$$
\forall B. ((A \to B) \to B)
$$
and then it is indeed the case (in classical and intutionistic logic) that the following sentence is provable:
$$
\forall A. (A \leftrightarrow (\forall B. ((A \to B) \to B)))
$$
(The left-to-right direction is trivial and the right-to-left direction follows by instantiating $B$ to $A$.)
What you refer to as the "dual" statement, restated in second-order logic:
$$
\forall A. (A \leftrightarrow (\exists B. ((B \to A) \land B)))
$$
is also provable (in classical or intuitionistic logic). (Right-to-left is now trivial and you can take $A$ as the witness going left-to-right.)
(If you'd like more information about the intuitionistic point of view on this please add a comment.)
