We consider h = Exp[g].
The main idea is to calculate the Fourier transform of h, expand it into a series in powers of v, and then transform back
Here we go.
The normal distribution is explicitly
f = 1/(v Sqrt[2 \[Pi]]) Exp[-(x - z)^2/(2 v^2)];
Now the Fourier transformation of h acts only on f, with the result
ft = FourierTransform[f, x, t, Assumptions -> v > 0]
(*
Out[49]= E^(-(1/2) t^2 v^2 + I t z)/Sqrt[2 \[Pi]]
*)
For simplicity let (with a > 0)
y[z_] := a z^2
so that the convolution becomes
cft = Integrate[
Exp[-y[z]] E^(-(1/2) t^2 v^2 + I t z)/Sqrt[
2 \[Pi]], {z, -\[Infinity], \[Infinity]}, Assumptions -> a > 0]
(*
Out[51]= E^(-((t^2 (1 + 2 a v^2))/(4 a)))/(Sqrt[2] Sqrt[a])
*)
Now we expand in powers of v
scft = (Series[cft, {v, 0, 4}] // Normal)
(*
Out[53]= E^(-(t^2/(4 a)))/(Sqrt[2] Sqrt[a]) - (E^(-(t^2/(4 a))) t^2 v^2)/(
2 Sqrt[2] Sqrt[a]) + (E^(-(t^2/(4 a))) t^4 v^4)/(8 Sqrt[2] Sqrt[a])
*)
and transform back
h = InverseFourierTransform[scft, t, x, Assumptions -> a > 0]
(*
Out[54]= 1/2 E^(-a x^2) (2 - 2 a v^2 - 12 a^3 v^4 x^2 + 4 a^4 v^4 x^4 +
a^2 (3 v^4 + 4 v^2 x^2))
*)
For general y[z] we don't calculate the z integral in the first place but write for the Fourier transform of h
hf = Integrate[
Exp[-y[z]] Exp[-(1/2) t^2 v^2 + I t z], {z, -\[Infinity], \[Infinity]}]
Now we see that the z-integral just gives the Fourier transform (Ft) of Exp[g] giving
hf = Ft[Exp[-y[z]] , z, t] Exp[-(1/2) t^2 v^2]
This must be transformed back (iFt), giving
h = iFt[ Ft[Exp[-y[z]] , z, t] Exp[-(1/2) t^2 v^2], t, x]
The expansion into powers of v is obvious, giving powers in t^2. The final g = Log[h] can then be expanded into powers of v, correspondingly.
I hope this is the general expression you were looking for.