= Solution
Choose the positive-$x$ reference <ground state> and use the cyclic <rotation>
$$
\widetilde S^z=S^x,\qquad\widetilde S^x=S^y,\qquad\widetilde S^y=S^z.
$$
This is a right-handed permutation, so it preserves the <spin commutation relations>. It puts the ordered direction along the <Holstein–Primakoff transformation>'s quantization axis. To the required order,
$$
\widetilde S^z_j=S-n_j,\quad
\widetilde S^x_j=\sqrt{\frac S2}(a_j+a_j^\dagger),\quad
\widetilde S^y_j=-i\sqrt{\frac S2}(a_j-a_j^\dagger).
$$
The longitudinal exchange contributes $-LJ^xS^2+2SJ^x\sum_jn_j+O(S^0)$. The transverse terms contribute
$$
-\frac S2\sum_j\left[(J^y+J^z)(a_j^\dagger a_{j+1}+a_{j+1}^\dagger a_j)
+(J^y-J^z)(a_ja_{j+1}+a_j^\dagger a_{j+1}^\dagger)\right].
$$
Dropping the constant and all terms of order $S^0$ or smaller gives the stated <linear spin-wave approximation>, with
$$
\boxed{A=-\frac S2(J^y+J^z),\qquad B=-\frac S2(J^y-J^z),\qquad C=2SJ^x.}
$$
The anomalous pair terms are present because the two transverse couplings differ. A different right-handed choice of transverse axes can change the sign of $B$ by a boson phase convention, while leaving the <dispersion relation> unchanged.
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