Why is the ROC of Laplace transform independent of imaginary part of s?
I think your misunderstanding is that you manipulate an expression, which is only valid for $\text{Re}\{s\}&gt;0$, in order to show for which values of $s$ it might be valid.

Note that the integral

$$\mathcal{L}\{u(t)\}=\int_0^{\infty}e^{-st}dt$$

only converges for $\text{Re}\{s\}&gt;0$, so the result $1/s$ does not make any sense for other values of $s$.

Only the real part of $s$ in the term $e^{-st}$ inside the integral can provide the necessary damping such that the integrand decays sufficiently fast for the integral to converge. That's why the region of convergence only depends on the real part of $s$.