NP by definition is the class of decision problems solvable in polynomial-time on a non-deterministic Turing machine. So, by definition, NP is your $A$ set. NP can't also be $B$ for superficial reasons: NP is a subset of functions $2^*\to 2$, and so it simply doesn't contain any functions $2^*\to2^*$ whose output isn't always one bit long.
There is a separate class FNP for the class of search problems solvable in polynomial-time on a non-deterministic Turing machine. A decision problem $\varphi$ is in NP when $$\forall x\in 2^*.\varphi(x)\iff \exists u\in2^{p(|x|)}.M(x,u)=1$$ for some polynomial $p$ and polynomial-time (deterministic) Turing machine $M$. A function $f$ is in FNP if given $x$ it finds a $u$ such that $M(x,u)=1$. Note, there is no constraint on how much time $f$ takes to produce a $u$ from $x$. That said, on a non-deterministic Turing machine, we can just guess the bits of $u$ and then check them against $M$ in polynomial-time. (We can cast an arbitrary function computed in polynomial-time on an NDTM $F$ as the relation $R(x,\langle u,y\rangle)\iff F(x,u)=y$ where $u$ represents the non-deterministic choices which is calculable in polynomial-time simply by simulating $F$ with choices $u$.)
As mentioned on the Complexity Zoo page, The Complexity of Decision Versus Search shows that if EE $\neq$ NEE or (citing a different result) if NE $\neq$ coNE then there are problems in FNP that can't be reduced to corresponding decision problems in NP in the sense if for any suitable machine $M$ we're provided an oracle for $\varphi(x)\equiv\exists u\in 2^{p(|x|)}.M(x,u)=1$ we could compute in polynomial-time some $u$ for the particular $x$ we were given such that $M(x,u)=1$. Any decision problem whose corresponding search problem is reducible to the decision problem can't be NP-complete: given an oracle for an NP-complete problem, we'd then be able to efficiently compute SAT whose search problem we can reduce to the corresponding decision problem, and which we can then use to compute the $u$ for the original problem by reducing it to SAT.
So the focus on decision problems doesn't seem completely innocuous, but we do have P = NP $\iff$ FP = FNP, so for at least the P = NP problem it doesn't matter. This can be partially seen from the above since if P = NP then every problem in NP is NP-complete and the counter-example to the reduction of search to decision doesn't exist.
