Let X denote a real Banach space, and consider then the ODE \begin{equation} \begin{cases} \mathbf{u}'(t)=A \mathbf{u}(t); \ \ \ (t\geq0)\\ \mathbf{u}(0)=u, \end{cases} \end{equation} where $'=\frac{d}{dt}, u\in X$ is given, and A is a linear operator. We assume that the ODE has a unique solution for each initial point and we will write $$\mathbf{u}(t):=S(t) u \ \ \ (t\geq0)$$ to display explicitly the dependence of $\mathbf{u}(t)$ on the initial value $u\in X$.
On my textbook some properties of S(t) are discussed. For example
$$S(t+s)u=S(t)S(s)u=S(s)S(t)u \ \ \ (t,s\geq 0, u\in X).$$
On textbook one can read: "this condition is simply our assumption that the ODE has unique solution for each initial point". I can not understand the link between the previous condition and the uniqueness of the solution. Any suggestions please?
Set $\mathbf u(t):=S(t)u$ as you did. Also, fix $s$ and define $$\mathbf v(t):=S(t+s)u\quad{\rm and}\quad\mathbf w(t):=S(t)S(s)u\, .$$
Then $\mathbf v(t)=\mathbf u(t+s)$, so that $$\mathbf v'(t)=\mathbf u'(t+s)=A\mathbf u(t+s)=A\mathbf v(t)\, ; $$ moreover, $\mathbf v(0)=S(s)u$.
On the other hand, $\mathbf w(t)$ is by definition the solution of the ODE with initial value $S(s)u$.
Hence, $\mathbf v=\mathbf w$ by uniqueness of the solution; which is the required result.