There is no reason to think it has a closed form.
One way is calculate numerically:
$Version
(*12.3.0 for Microsoft Windows (64-bit) (May 10, 2021)*)
f[x_, t_, F_, u0_, \[Lambda]_] :=
InverseLaplaceTransform[(
E^(-Sqrt[s] x + Sqrt[s] (2 + x))
u0 (Sqrt[s] - \[Lambda]))/((-1 + E^(2 Sqrt[s])) s^(
3/2) - (1 + E^(2 Sqrt[s])) s \[Lambda]) - (
u0 (Sqrt[s] + \[Lambda]))/((-1 + E^(2 Sqrt[s])) s^(
3/2) - (1 + E^(2 Sqrt[s])) s \[Lambda]) + (
E^(Sqrt[s] x) (-F + u0 \[Lambda]))/((-1 + E^(2 Sqrt[s])) s^(
3/2) - (1 + E^(2 Sqrt[s])) s \[Lambda]) + (
E^(2 Sqrt[s] -
Sqrt[s] x) (-F + u0 \[Lambda]))/((-1 + E^(2 Sqrt[s])) s^(
3/2) - (1 + E^(2 Sqrt[s])) s \[Lambda]), s, N@t,
Method -> "Crump"]
ListLinePlot[Table[{t, f[1/2, t, 1, 2, 3]}, {t, 0.001, 4, 1/30}],
PlotRange -> All](*for: x=1/2,F=1,u0=2,\[Lambda]=3*)
data = Flatten[
Table[{x, t, f[x, t, 1, 2, 3]}, {x, -1, 1, 1/5}, {t, 0.001, 3,
1/5}], 1];
ListPlot3D[data, InterpolationOrder -> 3,
AxesLabel -> {"x", "t"}](*for: F=1,u0=2,\[Lambda]=3*)