Dirichlet series for inverse of Eta function We know that $$ \frac{1}{\zeta (s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^{s}} $$
but what happens with  $$ \frac{1}{\eta (s)} = \sum_{n=1}^{\infty} \frac{b(n)}{n^{s}} $$
with $ \eta (s) =  (1-2^{1-s})\zeta (s) ?$
Can I evaluate the coefficients $ b(n) $ ? Perhaps I should apply the mobius transform to the sequence $ a(n)= (-1)^{n} .$

You are correct that $b(n)$ can be computed recursively as the Dirichlet inverse of $a(n)$. But there's another way: simply expand $(1-2^{1-s})^{-1}$ and multiply that against $\zeta(s)^{-1}$:
$$\frac{1}{\eta(s)}=\cfrac{1}{\displaystyle 1-\frac{2}{2^s}}\frac{1}{\zeta(s)}=\left(1+\frac{2}{2^s}+\frac{4}{4^s}+\frac{8}{8^s}+\cdots\right)\sum_{n=1}^\infty\frac{\mu(n)}{n^s}=\sum_{n=1}^\infty\frac{b(n)}{n^s} $$
Therefore, for $n$ odd, $b(n)=\mu(n)$, and for $n$ even,
$$\begin{array}{ll} b(n) & =\sum_{\substack{d\mid n \\ d=2^r}}d\mu(n/d) \\ & =\sum_{r=0}^{v}2^r\mu(n/2^r) \\ & =2^{v_2(n)}\mu(n/2^{v_2(n)})+2^{v_2(n)-1}\mu(n/2^{v_2(n)-1}) \\[3pt] & =2^{v_2(n)-1}\mu(n/2^{v_2(n)}).\end{array}$$
