I've made some progress which gives me some encouragement the inverse transform is convergent. The code below numerically integrates the Bromwich integral with all parameters equal to one including the initial condition x(0)=1 and since the imaginary component is odd and the real even (by just plotting them), I just multiply the real part by 2 (but the 2i cancel) and integrate from s=6+0i to s=6+500i (since the pole in the denominator is at about 4.9) for t in the range of (0,1). I then fit the data to a 10-degree polynomial and then back-substitute that polynomial into the IDE and then compare plots of the left and right side of the equation by superimposing them. The plot below shows at least in this range, good agreement.

Code:

u[s_] := (1 - Sqrt[Pi/s])/(s - 1 - Sqrt[Pi*s]);
myTable = Table[{tval, (1/Pi)*Re[NIntegrate[u[s]*Exp[s*t] /.
{t -> tval, s -> 6 + I*y}, {y, 0, 500}]]}, {tval, 0.01, 1, 0.05}];
g[t_] = Sum[Subscript[a, n]*t^n, {n, 0, 10}]
theCoeff = Table[Subscript[a, n], {n, 0, 10}]
mycoef = FindFit[myTable, g[t], theCoeff, t]
mysol[t_] := g[t] /. mycoef;
leftside = D[mysol[t], t]
rightside = mysol[t] + Integrate[D[mysol[r], r]*(1/Sqrt[t - r]), {r, 0, t},
Assumptions -> t > 0]
left = Plot[leftside, {t, 0, 1}, PlotStyle -> Blue]
right = Plot[rightside, {t, 0, 1}, PlotStyle -> Red]
Show[{left, right}]