= Solution
In character theory, a <p-elementary group> is a finite group isomorphic to $P\times C$, where $P$ is a $p$-group and $C$ is cyclic of order coprime to $p$. Either factor may be trivial. An <elementary group> is p-elementary for at least one prime $p$; it need not be an elementary <abelian group>.
A <generalised character> means an integer linear combination of irreducible complex characters. <Brauer's characterisation of characters> states that a complex <class function> $f$ on a finite group $G$ is a <generalised character> if and only if $f_E$ is a <generalised character> for every elementary subgroup $E\leq G$. Equivalently,
$$
\boxed{f\in\mathbb Z\operatorname{Irr}(G)\iff\langle f_E,\alpha\rangle_E\in\mathbb Z\text{ for every elementary }E\leq G\text{ and }\alpha\in\operatorname{Irr}(E).}
$$
An ordinary character additionally requires nonnegative coefficients in its global irreducible decomposition. We use the virtual-character criterion just stated.
For the extension construction, set $d=\theta(1)$. The <determinant character> $\det\theta$ is linear. Since its extension $\mu$ has the same value $1$ at the identity, $\mu$ is also a <linear character>. For any $g\in G$, the subgroup $J_g=N\langle g\rangle$ has cyclic quotient $J_g/N$, hence a solvable quotient. The invariant character $\theta$ has degree coprime to $[J_g:N]$, since this index divides $[G:N]$, and $\mu_{J_g}$ extends its determinant. The permitted <coprime-degree determinant extension theorem> therefore gives a unique character $\chi_g\in\operatorname{Irr}(J_g)$ with
$$
(\chi_g)_N=\theta,\qquad\det\chi_g=\mu_{J_g}.
$$
Define a function on $G$ by $f(g)=\chi_g(g)$. We next verify that these individually defined values form a compatible <class function>.
If $x\in G$, conjugation carries $J_g$ to $J_{xgx^{-1}}$. Transporting $\chi_g$ by this conjugation gives a character extending $\theta$, because $\theta$ is $G$-invariant, and with determinant $\mu$ restricted to the conjugate subgroup, because $\mu$ is a <class function>. Uniqueness identifies the transported character with $\chi_{xgx^{-1}}$. Consequently $f(xgx^{-1})=f(g)$.
Now take an elementary subgroup $E\leq G$ and let $J=NE$. A finite $p$-group is solvable, a cyclic group is solvable, and a direct product of <solvable groups> is solvable. Thus $E$ is solvable, and so is $J/N\cong E/(E\cap N)$. Again the index $[J:N]$ divides $[G:N]$, so the same extension theorem gives $\chi_J\in\operatorname{Irr}(J)$ extending $\theta$ with determinant $\mu_J$.
For $g\in E$, we have $J_g\leq J$. The restriction $(\chi_J)_{J_g}$ is irreducible: an <invariant subspace> for $J_g$ would be invariant for $N$, whose restriction already affords the irreducible $\theta$. Its determinant is $\mu_{J_g}$. By uniqueness on $J_g$, it equals $\chi_g$. Therefore
$$
f_E=(\chi_J)_E,
$$
an ordinary character and hence a <generalised character> of $E$. <Brauer's characterisation of characters> now makes $f$ a <generalised character> of $G$. If $n\in N$, then $J_n=N$ and $\chi_n=\theta$, so $f(n)=\theta(n)$. We have proved, by <gluing determinant-normalized character extensions>,
$$
\boxed{\chi:=f\text{ is a generalised character of }G,\qquad\chi_N=\theta.}
$$
Only the intermediate quotients $J_g/N$ and $NE/N$ were required to be solvable; no solvability assumption on $G/N$ has been added.
For the prime-set assertion, let $E=P\times C$ be p-elementary, and split the cyclic group as $C=C_\pi\times C_{\pi'}$, where each factor has the indicated prime divisors in its order. If $p\in\pi$, put $E_\pi=P\times C_\pi$ and $E_{\pi'}=C_{\pi'}$. If $p\notin\pi$, put $E_\pi=C_\pi$ and $E_{\pi'}=P\times C_{\pi'}$. In both cases the factors commute, have trivial intersection and have orders with disjoint prime spectra. This proves the <prime-set decomposition of elementary groups>
$$
\boxed{E=E_\pi\times E_{\pi'},\qquad E_\pi\text{ a }\pi\text{-group},\quad E_{\pi'}\text{ a }\pi'\text{-group}.}
$$
Here a <pi-group> has order divisible only by primes in $\pi$, and a <pi-element> has such an element order; the identity qualifies for both complementary prime sets.
Finally, suppose $K$ satisfies the given element-order condition. The sets $A,B$ are unions of <conjugacy classes> and are disjoint, because membership in both would force every prime divisor of the element order into $\pi\cap\pi'=\varnothing$, hence force order $1$. Let $E\leq K$ be elementary and use $E=E_\pi\times E_{\pi'}$. If both factors were nontrivial, choose $a\ne1$ in the first and $b\ne1$ in the second. They commute and have coprime orders, so $ab$ has order $|a||b|$, containing primes from both $\pi$ and $\pi'$. It would belong to neither $A$ nor $B$ nor $\{1\}$, contrary to the hypothesis. Thus every elementary subgroup is entirely a <pi-group> or entirely a complementary-prime group.
Write $m=|K|_\pi$ and $r=|K|_{\pi'}$ for the two prime parts of $|K|$, so $|K|=mr$ and $\gcd(m,r)=1$. The <Chinese remainder theorem> supplies an integer $D$ satisfying
$$
D\equiv1\pmod m,\qquad D\equiv0\pmod r.
$$
Define the <class function>
$$
f(1)=D,\qquad f(a)=1\ (a\in A),\qquad f(b)=0\ (b\in B).
$$
For an elementary $\pi$-subgroup $E$, its order divides $m$. Let $\rho_E$ be the character of its <regular representation>, equal to $|E|$ at $1$ and zero elsewhere. Then
$$
f_E=1_E+\frac{D-1}{|E|}\rho_E,
$$
with integral coefficient because $D\equiv1\pmod{|E|}$. For an elementary $\pi'$-subgroup, its order divides $r$, and
$$
f_E=\frac{D}{|E|}\rho_E
$$
again has integral coefficient. The trivial subgroup satisfies either formula. Every elementary restriction is therefore a <generalised character>. A second application of <Brauer's characterisation of characters> proves
$$
\boxed{\xi:=f\text{ is a generalised character of }K,\qquad\xi(a)=1\ (a\in A),\quad\xi(b)=0\ (b\in B).}
$$
The freely chosen identity value, fixed by the two congruences, makes the restrictions integral. This construction also covers the cases $\pi=\varnothing$, $\pi'=\varnothing$, or $K=\{1\}$, using modulus-one congruences where appropriate.
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