= Solution
Let $\mathbf V$ be the <centre of mass> velocity and $\mathbf v$ the binary relative velocity. Summing the two linear <gas drag> forces gives
$$
\dot{\mathbf V}\big|_{\rm drag}
=-\kappa_b(\mathbf V-\mathbf v_g)
-\frac{m_1m_2}{(m_1+m_2)^2}\Delta\mathbf v,
\qquad
\kappa_b=\frac{m_1/t_{s1}+m_2/t_{s2}}{m_1+m_2}.
$$
The last term averages to zero over the internal circular orbit. On the nearly circular stellar orbit, $\mathbf V\simeq v_K\hat{\boldsymbol\theta}$ and $\mathbf v_g=(1-\eta)v_K\hat{\boldsymbol\theta}$, so the averaged tangential acceleration is $-\eta\kappa_bv_K$.
With <specific orbital energy> $-GM_\star/(2a_b)$, the <orbit-averaged drag work> is $-\eta\kappa_bv_K^2$, where $v_K^2=GM_\star/a_b$. Thus <gas-driven migration of a binary centre of mass> gives
$$
\boxed{\dot a_b=-2\eta\kappa_ba_b=-\frac{a_b}{\tau_b},\qquad
\tau_b=\frac{m_1+m_2}{2\eta(m_1/t_{s1}+m_2/t_{s2})}.}
$$
This uses the weak-drag, nearly Keplerian approximation implicit in applying the circular-orbit work relation. Strongly gas-coupled orbits require a coupled radial-azimuthal drift solution.
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