Difference between the two forms of input-output relationships of an LTI system
For an LTI system with impulse response $h(t)$ and transfer function $H(s)=\mathcal{L}\{h(t)\}$, an input signal $x(t)=e^{st}$ (note: no multiplication with the unit step function $u(t)$), results in an output

$$\begin{align}y(t)&amp;=(x\star h)(t)\\&amp;=\int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau\\&amp;=\int_{-\infty}^{\infty}h(\tau)e^{s(t-\tau)}d\tau\\&amp;=e^{st}\int_{-\infty}^{\infty}h(\tau)e^{-s\tau}d\tau\\&amp;=e^{st}H(s)\tag{1}\end{align}$$

This is a very basic result concerning eigenfunctions of LTI systems.

If the input signal is given by $x(t)=e^{st}u(t)$ then the output is not given by $(1)$, but it needs to be determined from the convolution integral:

$$\begin{align}y(t)&amp;=\int_{-\infty}^{\infty}h(\tau)e^{s(t-\tau)}u(t-\tau)d\tau\\&amp;=\int_{-\infty}^th(\tau)e^{s(t-\tau)}d\tau\\&amp;=e^{st}\int_{-\infty}^th(\tau)e^{-s\tau}d\tau\end{align}$$

Note that in both cases you get the zero-state response because convolution with an impulse response cannot give you anything else.