Link between zero state response and the particular solution for an everlasting exponential input
I do not believe there is any problem with the notation used by the author. He just wanted to express the following (\$x\$ input, \$y\$ output - the response to a eigenfunction \$e^{st}\$):

$$y(t) = \int_{-\infty}^\infty h(\tau)x(t-\tau)d\tau  $$
$$ = \int_{-\infty}^\infty h(\tau)e^{s(t-\tau)}d\tau  $$
$$ = e^{st} \int_{-\infty}^\infty h(\tau)e^{-s\tau}d\tau  $$

Thus, the response to \$e^{st}\$ is of the form

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

where \$H(s)\$ is a complex constant whose value depends on \$s\$ and wich is related to the impulse response by

$$ H(s) = \int_{-\infty}^\infty h(\tau)e^{-s\tau}d\tau  $$

There is no conflict time x frequency here.

Also, note:

If \$\lambda\$ is a pole of \$H(s)\$, the mode \$e^{{\lambda}t}\$ can be generated at the output by an initial state without the application of any input. If \$\lambda\$ is not a pole of \$H(s)\$, then this is not possible. In this case, the only way to generate \$e^{{\lambda}t}\$ at the output is to apply \$e^{{\lambda}t}\$ at the input.  In other hand, if the input \$x(t)\$ is of the form \$e^{{\lambda}t}\$, where \$\lambda\$ is not a pole of \$H(s)\$, then the output due to input \$x(t) = e^{{\lambda}t}\$  is equal to \$y(t) = H(\lambda)e^{{\lambda}t}\$.