Here are three workarounds:

f[x_] := 1 - 1/(x - 9/2)
Show[Plot[f[x], {x, 0, 10}, Filling -> Axis, PlotStyle -> None,
  Ticks -> {Range[20], Automatic}, AxesOrigin -> {0, 0}],
 Plot[f[x], {x, 0, 10},
  ExclusionsStyle -> Directive[Dashed, AbsoluteThickness[0.5], Red], 
  Exclusions -> "Singularities"]]
Plot[f[x], {x, 0, 10}, Filling -> Axis, ExclusionsStyle -> None, 
 Exclusions -> "Singularities", 
 Epilog -> {Dashed, AbsoluteThickness[0.5], Red, 
   Map[InfiniteLine[#, {0, 1}] &, 
    SolveValues[{FunctionDiscontinuities[f[x], x], y == 0}, {x, y}]]}]
Plot[f[x], {x, 0, 10}, Filling -> Axis,
 ExclusionsStyle -> None, Exclusions -> "Singularities", 
 GridLines -> {SolveValues[FunctionDiscontinuities[f[x], x], x], None},
  GridLinesStyle -> Directive[Dashed, AbsoluteThickness[0.5], Red]]

An explicit Exclusions value changes the appearance of the asymptote:

f[x_] := 1 - 1/(x - 9/2);
Plot[f[x], {x, 0, 10}, Filling -> Axis, 
 Ticks -> {Range[20], Automatic}, AxesOrigin -> {0, 0}, 
 ExclusionsStyle -> Directive[Dashed, AbsoluteThickness[0.5], Red], 
 Exclusions -> 9/2]

Anyway, in this case I would recommend Exclusions -> "Singularities" instead of "Discontinuities".

It seems that Filling completely kills the Exclusions:

pl = Plot[Ceiling[x], {x, -5, 5},
  Filling -> Axis,
  ExclusionsStyle -> Red, Exclusions -> Range[-5, 5]]
FreeQ[pl, Red]

If you want both the effect of filling and the ability to freely change the style and color of the automatically generated function asymptote, how should the code be set?