Laplace of step and integration are same?
This is because the impulse response of an integrator is $h(t)=u(t)$. The output which is the convolution with the impulse respoponse is
$$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau$$
and with $h(t)=u(t)$ it becomes
$$\begin{align}
y(t)&amp;=\int_{-\infty}^{\infty}x(\tau)u(t-\tau)d\tau\\
&amp;=\int_{-\infty}^{t}x(\tau)d\tau\tag{1}\end{align}$$
where $(1)$ is resulted from the fact that $$u(t-\tau)=\begin{cases}0&amp; \forall \tau&gt;t\\
1 &amp; \text{otherwise}\end{cases}$$ 
The transfer function (which is the Laplace transform of impulse response) is $$H(s)=\mathcal{L}\{h(t)=u(t)\}=\frac{1}{s}$$