A $C^{0}$ semigroup has to do with the parameter of time in a time-dependent equation such as the heat equation
$$
\frac{\partial}{\partial t}h(x,t)=\Delta h(x,t),\\
h(x,0) = h_{0}(x)
$$
The solution operator $S(t)$ evolves the heat distribution $h_{0}$ at time $0$ to that at time $t$. That is $h(x,t)=S(t)h_{0}$. If the system is well posed, then there is only one solution at the later time. Furthermore, if you evolve $t$ seconds into the future and then use the new state $h(x,t) = S(t)h_{0}$ as an initial condition for evolving $t'$ more seconds, then you should get the same answer as if you were to evolve the initial state through $t+t'$ seconds. That is, $S(t')[S(t)h_{0}]=S(t+t')h_{0}$, which gives the solution operator an exponential property
$$
S(t')S(t) = S(t'+t).
$$
This exponential property is a general principle of time evolution systems where the system itself does not depend on time. The $C^{0}$ property has to do with stability of the solution. You want,
$$
\lim_{t\downarrow 0}h(x,t)=h_{0}, \mbox{ i.e., }\\
\lim_{t\downarrow 0}S(t)h_{0} = h_{0}.
$$
In other words, you want $S(0)=I$ with vector continuity $S(t)h_{0}\rightarrow Ih_{0}$ as $t\downarrow 0$. Continuity from above at $t=0$ gives the same for all later times, too.
If the system is time dependent, then the evolution operator requires the beginning and ending time, say $S(t_{\mbox{end}},t_{\mbox{start}})$. You still get something like an exponential property, but not quite as simple
$$
S(t_{3},t_{2})S(t_{2},t_{1})=S(t_{3},t_{1}).
$$
It is the interesting that time evolution for time-independent linear systems has this exponential property. It is an odd fact of nature. And the whole idea of $C^{0}$ semigroup theory is to try exploit this exponential property in order to write the solution operator as some kind of actual exponential operator $e^{tA}$ for an operator $A$. You can identify what $A$ would have to be in order have an exponential property. For example, in the case of the heat equation, that operator $A$ must be $\Delta$ because you expect
$$
\frac{d}{dt}(e^{tA}h) = A(e^{tA}h)
$$
Schrodinger's wave equation for a single non-relativistic particle is another example
$$
i\hbar\frac{\partial}{\partial t}\psi(x,t) = \left[-\frac{\hbar^{2}}{2\mu}\Delta+V\right]\psi(x,t),\\ \psi(x,0)=\psi_{0}(x).
$$
If the potential $V$ does not depend on time, then you expect a solution operator $e^{tA}$ with
$$
A = \frac{1}{i\hbar}\left[-\frac{\hbar^{2}}{2\mu}\Delta+V\right]
$$
In this case you expect $|\psi|^{2}$ to be a spatial probability distribution for each $t$, so that $\|e^{tA}\phi_{0}\|=\|\phi_{0}\|=1$ for all $t$. This corresponds to the case where $A=iH$ where $H=H^{\star}$ so that $e^{itH}$ is unitary for all $t$.
