why would a solution with H+ concentration of 1.00 x 10^-7 M be said to be neutral?

A solution is said to be neutral when [H+] = [OH-]
The ionic product of water Kw = 1*10^-14 at 25°C
Ionic product formula:
[H+] [OH-] = 1*10^-14 at 25°C
If [H+] = 1*10^-7M , solve for [OH-]
[OH-] = (1*10^-14) / (1*10^-7)
[OH-] = 1*10^-7M
Because [H+] = [OH-] the solution is neutral .
And you can calculate the pH of this solution:
pH = -log [H+]
pH = -log 1*10^-7
pH = 7.00
You can also calculate the pOH of the solution in the same way:
You get a result: pOH = 7.00
This is all at 25°C

What will happen if the temperature is increased to 50°C ?
The dissociation of water is an endothermic reaction. More water will dissociate to H+ and OH- ions
At 50°C , Kw = 5.476*10^-14
What is [H+] at 50°C
If the solution is neutral, and we have to accept that water is neutral,
Then [H+] =[OH-]
[H+] [OH-] = 5.476*10^-14
[H+] = √(5.476*10^-14)
[H+] 2.34*10^-7M
pH = 6.63

In the same way you can calculate [OH-] and pOH. You will find that [OH-] = 2.34*10^-7M and that pOH = 6.63.
At 50°C pure water has a neutral pH = 6.63. This water is not acidic, it is neutral because [H+] = [OH-]

pH is found by -log(H+). If you -log(1.00 x 10^-7) you'll get a value of 7 which is neutral.

Scientists have found that the concentration of H⁺ of any neutral solution at 25°C is 1 × 10⁻⁷ M. Besides, the concentration of OH⁻ of any neutral solution at 25°C is also 1 × 10⁻⁷ M.

Therefore, we can use the FACT that "a solution with H⁺ concentration of 1.00 × 10⁻⁷ M at 25°C is said to be neutral."