How can we prove the correctness of the integration property of the Laplace transform?
You are right that the argument in the proof is not correct, or at least misleading. The fact that $e^{-st}$ becomes zero as $t\to\infty$ is true for any $s$ with $\textrm{Re}\{s\}&gt;0$. However, the integral
$$\int_0^{\infty}f(\tau)d\tau$$
might not exist, so we can't just compute the limit by claiming that the first term becomes zero.
We have to compute
$$\lim_{t\to\infty}-\frac{e^{-st}}{s}\int_0^{t}f(\tau)d\tau\tag{1}$$
The limit $(1)$ only equals zero for $s$ inside the region of convergence (ROC) of $f(t)$, i.e., we don't just require that $e^{-st}\to 0$ for $t\to\infty$, but we require it to decay sufficiently fast to compensate for the growth of the integral.