Why is there no standard temperature?
The short answer

The standard pressure is not an experimental condition, while the temperature is.

The long answer

The main reason the standard state is defined is because it leads to this very useful equation
$$K_p = e^{-\Delta G^\circ/RT}$$
Say you have this reaction: $A \rightleftharpoons B + C$ One way to use this equation is to compute the free energy of 1 mol of $A$, $B$, and $C$ at 1 bar using equations derived for an ideal gas, compute $\Delta G^\circ = G^\circ (B) + G^\circ (C) - G^\circ (A)$, and use that value to predict $K_p$.

If the gasses behave like ideal gasses "in real life" then the measured  $K_p$ will match the $K_p$ computed from $\Delta G^\circ$.  You can do the measurement at any pressure you want, not just at 1 bar.* The standard state refers to the pressure you use when computing $\Delta G^\circ$.  The only thing it has to do with the experimental measurement is that it defines the units you should use for your partial pressures when computing $K_p$

$\Delta G^\circ$ does also depend on temperature, but the temperature you chose should be the same as the experimental conditions.  So the temperature is not part of the standard state definition.

But what about "Standard temperature and pressure (STP)?"

Standard temperature and pressure (STP) refers to STP conditions under which $K_p$ is measured, not the pressure used to compute $\Delta G^\circ$. I know, they couldn't have made it more confusing if they tried when they named these things.

*Of course if you do the measurements at very high pressures or low temperatures, then the assumption that the gasses behave ideally will be less valid and the measured $K_p$ will differ more from the $K_p$ computed from Equation 1.  However, that is a separate issue unrelated to the standard state because the $K_p$ in Equation 1 refers to the $K_p$ you would measure if the gasses behaved ideally at the pressure and temperature used in the experiment.