= Solution
Use the magnetic-coupling convention of the supplied <Hamiltonian>; electromagnetic unit factors are absorbed into the stated $e$ and $B$. Let $\alpha=m_1/M$ and $\beta=m_2/M$, so
$$
\mathbf r_1=\mathbf R-\beta\mathbf r,\qquad
\mathbf r_2=\mathbf R+\alpha\mathbf r.
$$
The <kinetic energy> separates into $M\dot{\mathbf R}^2/2+\mu\dot{\mathbf r}^2/2$. The <symmetric gauge> coupling $\sum_i e_i\dot{\mathbf r}_i\cdot(\mathbf B\times\mathbf r_i)/2$ becomes
$$
-\frac e2[\dot{\mathbf R}\cdot(\mathbf B\times\mathbf r)
+\dot{\mathbf r}\cdot(\mathbf B\times\mathbf R)]
-e^*\dot{\mathbf r}\cdot(\mathbf B\times\mathbf r),\quad
e^*=\frac{e(m_1-m_2)}{2M}.
$$
The <scalar triple product> identity gives
$$
\dot{\mathbf r}\cdot(\mathbf B\times\mathbf R)
=\frac d{dt}[\mathbf r\cdot(\mathbf B\times\mathbf R)]
+\dot{\mathbf R}\cdot(\mathbf B\times\mathbf r).
$$
Discarding $-e/2$ times that <total derivative>, which changes endpoint terms and <canonical momenta> but not the trajectories, leaves
$$
\boxed{L=\frac M2\dot{\mathbf R}^2+\frac\mu2\dot{\mathbf r}^2+\frac{e^2}{r}
-e^*\dot{\mathbf r}\cdot(\mathbf B\times\mathbf r)
-e\dot{\mathbf R}\cdot(\mathbf B\times\mathbf r).}
$$
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