Settling time is usually defined as when when the step response is within 2% of the steady state response. Even though the difference between the step response and the steady state response gets bigger for all times when you use a bigger gain $K$, the percentage difference remains the same. That is why $K$ does not affect settling time.
The poles of your system are,
$$
p = \omega_n \left(-\zeta \pm \sqrt{\zeta^2 - 1}\right).
$$
The transient response of such a system will look like,
$$
y_t(t) = C_1\, e^{\omega_n \left(-\zeta + \sqrt{\zeta^2 - 1}\right) t} + C_2\, e^{\omega_n \left(-\zeta - \sqrt{\zeta^2 - 1}\right) t},
$$
where $C_1$ and $C_2$ depend on the initial conditions.
For $\zeta<1$ the poles will be each others complex conjugate and the transient can also be written as,
$$
y_t(t) = e^{-\omega_n\, \zeta\, t} \left(C_3 \cos\left(\omega_n\,\sqrt{1 - \zeta^2}\, t\right) + C_4 \sin\left(\omega_n\,\sqrt{1 - \zeta^2}\, t\right)\right).
$$
For a step response the total response can be written as the sum of the steady state and the transient response ($y(t)=y_{ss}(t)+y_t(t)$), with the steady state equal to $K$ (which can be found by setting $s$ to zero). For a step response it is assumed that the system is initial at rest, so position and velocity are zero for all $t\leq0$. Applying these constraints yields $C_3=-K$ and $C_4=0$,
$$
y(t) = K \left(1 - e^{-\omega_n\, \zeta\, t} \cos\left(\omega_n\,\sqrt{1 - \zeta^2}\, t\right)\right).
$$
Using an upper bound of one for the cosine, then the settling time can be approximated by only considering the exponential,
$$
\left|\frac{y(t)-y_{ss}(t)}{y_{ss}(t)}\right| \approx e^{-\omega_n\, \zeta\, t}.
$$
It can be noted that if you work out the left hand side of this equation yourself you will see that all $K$'s cancel.
The settling time for this approximation can be found by solving when $e^{-\omega_n\, \zeta\, t}=0.02$. This has the solution,
$$
T_{settling} = \frac{-\ln(0.02)}{\zeta\,\omega_n},
$$
$-\ln(0.02)$ is indeed close to $4$.
Solving for the constants for two real poles (when $\zeta\geq 1$) you get,
$$
C_1 = K \frac{\sqrt{\zeta^2 - 1} + \zeta}{2 \sqrt{\zeta^2 - 1}},
$$
$$
C_2 = K \frac{\sqrt{\zeta^2 - 1} - \zeta}{2 \sqrt{\zeta^2 - 1}}.
$$
For $\zeta$ equal or slightly bigger than 1 you will have to solve the following equation for $t$ in order to get the settling time,
$$
\left|e^{-\omega_n\, \zeta\, t} \left(\cosh\left(\omega_n \sqrt{\zeta^2 - 1}\, t\right) + \frac{\zeta}{\sqrt{\zeta^2 - 1}} \sinh\left(\omega_n \sqrt{\zeta^2 - 1}\, t\right)\right)\right| = 0.02.
$$
When $\zeta\gg1$ then $\zeta\, (\zeta^2 - 1)^{-1/2} \approx 1$ and $\cosh(t)+\sinh(t)=e^t$ can be used. Solving the above equation then has the solution,
$$
T_{settling} = \frac{-\ln(0.02)}{\left(\zeta - \sqrt{\zeta^2 - 1}\right) \omega_n}.
$$
This can also be seen as taking the biggest (least negative) pole, instead of the real part of the complex conjugate poles. Because when $\zeta$ is large the difference between the two real poles is large and only the "slowest" pole will dominate the transient response for the settling time.
