Difference between the two forms of input-output relationships of an LTI system $$y(t)=f(t)*h(t)\tag{1}$$

$$y(t)=H(s)e^{st}\tag{2}$$

$$H(s)=\int_{-\infty}^{\infty}h(t)e^{-st}dt\tag{3}$$

Let $f(t)$ in Eqn $(1)$ be $e^{st}$. In many worked out examples, I have found that the two equations/formulae give different responses. To be specific, the actual convolution (Eqn $(1)$) is giving the zero-state response, while the Eqn $(2)$ involving $H(s)$ is giving the forced response. Recognising that Forced response and Zero-State response are not equal (Z-S responses may involve characteristic modes of the system, while the Forced Response is the non-characteristic mode part of the Total response), I am confused what is the intuitive difference between the two equations? Which equation should be primarily used for finding the response to an input to a system? Note that in the examples, the inputs $e^{st}$ were multiplied by $u(t)$.
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.