A partial answer is that the class includes all arithmetical sets, as follows. Suppose $S$ is defined by a formula $\psi(n) \equiv (\exists a)(\forall b)(\exists c)(\forall d) \cdots \phi(n, a, b, c, d, ...)$ where $\phi$ is quantifier free. By adding dummy quantifiers we can require that the quantifiers all occur in blocks of two like that, exists followed by forall. A function to decide $S$ from competing provers is as follows. First it asks $Y$ for a value of $a$. Then it asks $N$ for a value of $b$ given $a$. Then it asks $Y$ for a value of $c$ given $a$ and $b$, and so on. Once it has values for all the variables in $\phi$, it checks whether $\phi$ holds with those values plugged in. If so, it accepts $n$, and if not it rejects $n$.
If $n \in S$ then $Y$ has a winning strategy, because $\psi(n)$ is true. All $Y$ has to do is pick appropriate values for each existential quantifier, which have to exist if $\psi(n)$ holds. Otherwise, $\lnot \psi(n)$ is true, so $N$ has a winning strategy using the same technique of picking witnesses for the existential quantifiers in $\lnot \psi$.
Not every set in $S$ is arithmetical, though. As usual let $0^{a}$ be the $a$-th iterated Turing jump of the empty set. Consider $$S = 0^{(\omega)} = \{ 2^a3^b : b \in 0^{a}\}.$$ This is not an arithmetical set, but it can be accepted from competing provers as follows. First the machine looks at the input $n$ and decodes it into $2^a3^b$. If the number is not of that form the machine can just reject out of hand. If it is of that form, the machine pretends that it was given input $b$ and checks whether that number is in $0^{a}$ using $Y$ and $N$. Because an arithmetical formula for $0^{a}$ is uniformly recoverable from $a$, this can be done effectively, and $Y$ and $N$ will have the appropriate strategies because $0^{a}$ is arithmetical.
Note all the machines in this answer so far will halt on all strategies, and in fact we can give a bound on how many queries the machine will make as a function of $n$.