We can express Log[s+a] by series:
SeriesCoefficient[Log[s + a], {a, 0, n}]
We can check:
FullSimplify[Log[s] + Sum[((-1)^(1 + n) s^-n)/n*a^n, {n, 1, Infinity}], Assumptions -> {a > 0, s > 0}]
(*Log[a + s]*)
Then InverseLaplaceTransform is:
InverseLaplaceTransform[(Log[s] + ((-1)^(1 + n) s^-n)/n*a^n)/Sqrt[s], s, t]
(*((-1)^(1 + n) a^n t^(-(1/2) + n))/(
n Gamma[1/2 + n]) + (-Log[t] + PolyGamma[0, 1/2])/(
Sqrt[\[Pi]] Sqrt[t])*)
and
Sum[((-1)^(1 + n) a^n t^(-(1/2) + n))/(
n Gamma[1/2 + n]), {n, 1, Infinity}] + (-Log[t] +
PolyGamma[0, 1/2])/(Sqrt[\[Pi]] Sqrt[t])
(*(2 a Sqrt[t]
HypergeometricPFQ[{1, 1}, {3/2, 2}, -a t])/Sqrt[\[Pi]] + (-Log[t] +
PolyGamma[0, 1/2])/(Sqrt[\[Pi]] Sqrt[t])*)
$$\mathcal{L}_s^{-1}\left[\frac{\log (s+a)}{\sqrt{s}}\right](t)=\frac{2 a t \,
_2F_2\left(1,1;\frac{3}{2},2;-a t\right)-\log (t)+\psi
^{(0)}\left(\frac{1}{2}\right)}{\sqrt{\pi } \sqrt{t}}$$
Regards M.I.