Semigroup implies exponential Let $S :[0,1]\to \mathbb R^{n\times n}$ be a continuous function which satisfies
$$
S(0)=I \quad\text{and}\quad S(s+t)=S(s)S(t),
$$
for all $s,t\in[0,1]$ with $s+t\in[0,1]$. (Here $I$ is the identity matrix in $\mathbb R^{n\times n}$).
Show that there exists a matrix $A\in\mathbb R^{n\times n}$, such that $S(t)=\mathrm{e}^{tA}$.
Comment. Once we show that $S$ is differentiable at $0$, then setting $A=S'(0)$, it is not hard to prove the above, as we would then have $S'(t)=AS(t),\,\,S(0)=I$, which is a system of ODEs with unique solution the exponential.
 A: For $0<t<1$, define the integrated matrix   $H(t)x= \int_0^t S(w)x\,dw$.
Note that ${1\over t}H(t) x\to x$ as $t\to 0$, so for $t$ small enough, $H(t)$ is invertible. 
Now fix such a $t$.
Your semigroup is differentiable from the right on $H(t)(\mathbb{R}^n)=\mathbb{R}^n$ 
since direct calculations give
$${S(s)(H(t)x)-H(t)x\over s}\to (S(t)-I)x,$$ as $s\downarrow 0$. The required matrix is 
$A=(S(t)-I)H(t)^{-1}$. 
A: This may not be true; I will put my first impression and think about it. On a small neighborhood of the identity matrix we have a well-defined logarithm. So, taking $L(t) = \log S(t)$ we have the equation $L(s+t) = L(s) + L(t).$ This is Cauchy's functional equation in each matrix entry.
The interesting thing is that a bad solution of Cauchy's  equation is unbounded on any interval. Maybe you can conclude that $L$ is linear because of that.
Nope: take a discontinuous solution in one variable, 
$$ g(s+t) = g(s) + g(t). $$
Take a fixed matrix you like, call it $B.$ 
Then define 
$$ S(t) = e^{g(t) B}.  $$
This solves $S(s+t) = S(s) S(t).$ But kind of bad. So the question becomes, does this violate any of your hypotheses?  Hmmm. Yes, I think it makes for a discontinuous map $S.$ I will leave this here, I think it is instructive...
A: This is an elementary proof that given A such that $S(1)=e^A$ - i.e. if $S(1)$ admits a logarithm, then $S(t)=e^{t A}$.
You can see from your definition that $S(1)=$ $S(m (1/m))=$ $S(1/m + ... + 1/m)=$ $S(1/m)...S(1/m)=$ $S(1/m)^m$, so $e^A=S(1/m)^m$ and therefore $S(1/m)=e^{A{1/m}}$. Therefore, $S(n/m)=e^{(n/m)A}$ by similar argument for all $0\le n/m \le 1$. Note that that case $n=0$ requires $S(0)=I$ as per hypothesis.
You have two continuous functions, $S$ and $e^{t A}$, that are equal on $\Bbb Q$, which is dense in $[0, 1]$. Therefore, they are equal on the whole interval.
Update
I like Schmuland's answer, and won't elaborate further.
