Examples of functions that fulfill this property: $f(a + b) = f(a)f(b)$ for $a,b \in \mathbb{R}$. I'm not able to think of a function that works for non-whole numbers, though $f(x) = x^2$ works on $a=b \in \mathbb{N}/\{1\}$. Admittedly, not very close at all. This property is used in a proof I am attempting. 
Any ideas?
 A: First of all, we know that $f(0)=f(0+0)=f(0)^2$, so either $f(0)=0$ or $f(0)=1$. Moreover $f(a+0)=f(a)f(0)$, so if $f(0)=0$, then $f(a)=0$, for all $a$.
The constant function $f(x)=0$ satisfies the requirement.
Now, assume $f$ is not constant, so $f(0)=1$. Consider
$$
f(a)=f(a/2+a/2)=f(a/2)^2
$$
so $f(a)\ge0$. But, if $f(a)=0$, then $f(x)=f(a-a+x)=f(a)f(x-a)=0$ for all $x$, contrary to the hypothesis that $f$ is not constant. Therefore $f(a)>0$ for all $a$. If we consider the function
$$
g(x)=\log f(x)
$$
we have that
$$
g(a+b)=g(a)+g(b).
$$
This is a well known problem: $g$ is a $\mathbb{Q}$-linear function on $\mathbb{R}$. If you assume that $f$ (and so $g$) is continuous, then $g(x)=kx$ for some $k$ and so
$$
f(x)=k^x
$$
Assuming the axiom of choice, there are non continuous $\mathbb{Q}$-linear functions over $\mathbb{R}$. Each one of them gives a solution to your problem.
A: Look at $f(x) = e^x:$ $$f(a+b) = e^{a + b} = e^a\cdot e^b = f(a)f(b)$$
A: Answering my own questions more generally: 
Consider $f(x) = a^x$ where $a$ is a real number.
