The inverse series doesn't have that form. If $z=e^{-Q} (1+Q^{-1})$ then

$$Q = - \log z + \log (1+Q^{-1}) = \log z + (\log z)^{-1} + O((\log z)^{-2}) \quad \mbox{as} \ z \to 0^{+}.$$

The general form should be

$$Q = - \log z + \sum_{i>0} b_i (\log z)^i.$$

You might be able to coerce this into the Lagrange inversion form, but I don't see how right now. I would just generalize the solution above:

Write $$Q = - \log z + \log \left( 1 + \sum a_i Q^{-i} \right) = - \log z + \sum \frac{(-1)^k}{k} \left( \sum a_i Q^{-i} \right)^k.$$

Expanding this will give you a formula of the form $$Q = - \log z + \sum_{i=1}^N c_i Q^{-i} + O(Q^{-N-1}) \quad (*)$$ for any $N$ you like. If you already know that $Q = - \log z + \sum_{i=0}^{N-1} b_i (\log z)^{-i} + O((\log z)^{-N})$, then plug your known values into $(*)$ to deduce the value of $b_N$.

The inverse series doesn't have that form. If $z=e^{-Q} (1+Q^{-1})$ then

$$Q = - \log z + \log (1+Q^{-1}) = \log z + (\log z)^{-1} + O((\log z)^{-2}) \quad \mbox{as} \ z \to 0^{+}.$$

The general form should be

$$Q = - \log z + \sum_{i>0} b_i (\log z)^{-i}.$$

You might be able to coerce this into the Lagrange inversion form, but I don't see how right now. I would just generalize the solution above:

Write $$Q = - \log z + \log \left( 1 + \sum a_i Q^{-i} \right) = - \log z + \sum \frac{(-1)^k}{k} \left( \sum a_i Q^{-i} \right)^k.$$

Expanding this will give you a formula of the form $$Q = - \log z + \sum_{i=1}^N c_i Q^{-i} + O(Q^{-N-1}) \quad (*)$$ for any $N$ you like. If you already know that $Q = - \log z + \sum_{i=0}^{N-1} b_i (\log z)^{-i} + O((\log z)^{-N})$, then plug your known values into $(*)$ to deduce the value of $b_N$.