= Solution
If $A$ is empty, take the zero <generalized character>; if $B$ is empty, take the <trivial character>. Otherwise let $\pi$ be the set of primes dividing the orders of elements of $A$. The coprimality hypothesis makes every element of $B$ a $\pi'$-element. Thus every nonidentity element of $G$ is either a <pi-element> in $A$ or a complementary-prime element in $B$. In particular these sets are unions of <conjugacy classes>, since conjugation preserves element order.
Put $m=|G|_\pi$ and $n=|G|_{\pi'}$. Their <greatest common divisor> is one, so the <Chinese remainder theorem> supplies an integer $d$ such that
$$
d\equiv1\pmod m,\qquad d\equiv0\pmod n.
$$
Define the <class function> $f$ by $f(1)=d$, $f=1$ on $A$ and $f=0$ on $B$. We verify <Brauer's characterization of characters> directly.
For an <elementary subgroup> $E=P\times C$, the cyclic factor can be split into its $\pi$-part and $\pi'$-part. Attach $P$ to the part containing its prime. This expresses $E=E_\pi\times E_{\pi'}$. Both factors cannot be nontrivial: choosing a nonidentity element in each would give a commuting product of mixed prime order, whereas every element of $G$ has order involving only one of the two prime sets. Hence $E$ is wholly a $\pi$-group or wholly a $\pi'$-group.
Let $r_E$ be the <character> of the <regular representation> of $E$, with value $|E|$ at the identity and zero elsewhere. In the two cases respectively,
$$
f_E=1_E+\frac{d-1}{|E|}r_E,\qquad
f_E=\frac d{|E|}r_E.
$$
The coefficients are integers because $|E|$ divides $m$ or $n$. Thus both are <generalized characters>. The trivial <subgroup> causes no difficulty since $d$ is an integer. By <Brauer's characterization of characters>, $f$ is a <generalized character> of $G$.
\b[Taking $\vartheta=f$ gives the required values.]
This is a <generalized character separating complementary prime elements>; the freely chosen value at the identity is what makes all elementary restrictions integral.
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