If you do the transformation for each coefficient separately, Mathematica seems to work with Apart[].
Take
f[s_] :=(((1 + s)^2 + 3*s^(3/2)*(1 + s)^(1/2)*th - 4*s*(1 + s)*th + 3*s^(1/2)*(1 + s)^(3/2)*th + s^2*th^2))/(8 (s^(3/2)*(1 + s)^(1/2)*((1 + s)^(1/2) + s^(1/2)*th)^3));
fCoeff = Table[Expand[SeriesCoefficient[f[s], {th, 0, k}]], {k, 0, 40, 1}];
which is a list of 41 coefficients with respect to "th". Then
InverseLaplaceTransform[Apart[fCoeff[[k]]], s, t]
gives the k-th coefficient.
Honestly, the computation time for some coefficients is quite long. So if you are brave, you can do something like
Table[InverseLaplaceTransform[Apart[fCoeff[[k]]], s, t], {k, 1, 10}]
which still works pretty well on my computer. This list contains the coefficients with respect to your variable th.