Note that you only get a transient if you switch the system (or the input) on at a certain finite point in time. If the input $x(t)=e^{s_0t}$ has existed forever, the output is given by the convolution integral

\begin{align*} y(t)&=\int_{-\infty}^{\infty}h(\tau)e^{s_0(t-\tau)}d\tau\\&=e^{s_0t}\int_{-\infty}^{\infty}h(\tau)e^{-s_0\tau}d\tau\\&=e^{s_0t}H(s_0)\tag{1} \end{align*}

where $h(t)$ is the system's impulse response, and $H(s)=\mathcal{L}\{h(t)\}$ is the transfer function. Of course, in $(1)$ we've assumed that $s_0$ lies in the region of convergence of $H(s)$.

Note that you only get a transient if you switch the system (or the input) on at a certain finite point in time. If the input $x(t)=e^{s_0t}$ has existed forever, the output is given by the convolution integral

\begin{align*} y(t)&=\int_{-\infty}^{\infty}h(\tau)e^{s_0(t-\tau)}d\tau\\&=e^{s_0t}\int_{-\infty}^{\infty}h(\tau)e^{-s_0\tau}d\tau\\&=e^{s_0t}H(s_0)\tag{1} \end{align*}

where $h(t)$ is the system's impulse response, and $H(s)=\mathcal{L}\{h(t)\}$ is the transfer function. Of course, in $(1)$ we've assumed that $s_0$ lies in the region of convergence of $H(s)$. From $(1)$ we can see that $e^{s_0t}$ is an eigenfunction with eigenvalue $H(s_0)$.