Suppose $N(t)$ be a family of bounded $n$-by$n$ matrices with $t>0$ such that
$N(s+h)=N(s)N(h)$ holds for all $s,h>0$
What kind of structure of the solution could enjoy?
Can one always have that $N(t)=e^{At}$ holds for some $A$?
This assumption seems too strong, for instance, $N(s)=0$ is some solution.
So what structure does the whole solutions enjoy?
Maybe you are already aware of this, but in response to the setup that you have, the following comes to mind: One parameter semigroups
The linked PDF discusses evolution equations of the form $f(s+t)=f(s)f(t)$ in great detail, and might prove helpful.
As Thierry and Qiaochu have pointed out, it's not clear what exactly the question means. Even so, maybe it's possible to say something helpful.
The question asks
Can one always have that $N(t) = M^t$ holds for some non-singular $M$?
This doesn't quite make sense, so I'll interpret it as
Must there exist a non-singular matrix $M$ such that $N(t) = M^t$ for all $t$?
I'm guessing this was the intention.
The answer is probably no. (I say "probably" because the hypotheses on $N$ weren't stated precisely.) It's even false for $n = 1$. That is, there exist functions $N: \mathbb{R} \to (0, \infty)$ satisfying $$ N(s + t) = N(s) N(t) $$ for all $s, t \in \mathbb{R}$, but not of the form $N(t) = M^t$ for any $M > 0$. The existence of such functions requires the axiom of choice.
This is a problem going back to Cauchy. It's equivalent to consider the equation $f(s + t) = f(s) + f(t)$ (by putting $f(s) = \log N(s)$), and this is what's usually called "Cauchy's functional equation". You can find information about it here and here, for example.
Maybe you're happy to assume that $N(t)$ is continuous in $t$, in which case there's a genuine question about matrices to be answered. But in the absence of precise hypotheses, I don't know what you're happy to assume.
Let $N$ be the solution to a linear differential equation $\dot{N}(t) = A(t) N(t)$, with appropriate initial conditions. Then $N$ has the property you desire, but it cannot be represented as an exponential unless $A$ is independent of $t$. You can find this and more in books on linear system theory such as Jack Rugh's (http://books.google.com/books/about/Linear_system_theory.html?id=ffNQAAAAMAAJ).
I am not sure but I believe it is true that the converse also holds - under suitable technical conditions, if $N$ has the one-parameter semigroup property then it can be represented as the solution to a linear differential equation.
I shall assume $N(s)$ depends continuously on $s$; otherwise there are many discontinuous solutions even in the one-dimensional case. After a change of basis, we may assume that $$N(1)=\pmatrix{M&0\cr 0&Q},$$ where $M$ is nonsingular and $Q$ is nilpotent. Since all the matrices commute, we have the same block structure for all of them: $$N(s)=\pmatrix{M(s)&0\cr 0&Q(s)}.$$
Since $Q(1)$ is nilpotent, and $Q(1)=Q(1/n)^n$, it is easy to see that $Q$ is nilpotent for every rational. This is possible only if $Q(s)$ is identically zero. Moreover, $M$ is always nonsingular: if $s<1$, then $M(1)=M(s)M(1-s)$, so $M(s)$ is nonsingular for $s<1$, and for $s>1$, we can use that $M(s)=M(s/n)^n$. Now continuity implies that $M(\epsilon)=M(1)^{-1}M(1+\epsilon)$ approaches the identity as $\epsilon\to 0$, and it follows that $M(s)$ must be of the form $\exp(As)$.
