Why does spontaneity of a reaction depend on temperature?

The issue here is in your definition of entropy. According to your definition:

$$mathrm dS = frac{text{đ}q}{T}.tag{1}$$

However, there is a small difference.

As you may know, the entropy of a system is a state function and so it only depends on the initial and final states. Therefore, if the entropy of the system were to be calculated for a specific path for which the parameters are simple to find, then the value would not change irrespective of the path.

The actual formula of change in entropy is:

$$mathrm dS = frac{text{đ}q_mathrm{rev}}{T}.tag{2}$$

This value is calculated for a reversible isothermal process.

Now, let us assume that we have taken such an isothermal reversible process and discovered the value of $Delta S$ and now used an isobaric process to discover the value of $Delta H.$ Now, we solve for $Delta G$ in our isothermal process (this is done so that we can assume constant temperature in our formula).

Therefore, for such a process

$$q_mathrm{rev} = nRTlnfrac{V_ce{B}}{V_ce{A}}.tag{3}$$

Thus, we get change in entropy as

$$Delta S = nRlnfrac{V_ce{B}}{V_ce{A}}.tag{4}$$

Now, enthalpy would be a constant value that we discerned from the isobaric process.

Therefore, since $Delta G = Delta H - TDelta S,$ we get

$$Delta G = Delta H - nRTlnfrac{V_ce{B}}{V_ce{A}}.tag{5}$$

As you can see, there is still a temperature dependence.

This is because $Delta S$ is temperature independent and so you would still have a temperature dependence for $Delta G.$

Another thing that you may have forgotten to take into consideration is the fact that $q$ is dependent on the temperature.

The above example is that for an ideal gas expansion.

As Poutnik stated in the comments in case of a reaction, the calculation of $Delta H$ becomes more complicated as we have to take into consideration the different molar capacities of the reagents and products which would $Delta H$ would also be a function of temperature.

For standard value of entropy, we use the formula

$$Delta S^circ_mathrm{rxn} = sum nS^circ_mathrm{products} - sum nS^circ_mathrm{reactants}.tag{6}$$