A function in which addition and multiplication behave the same way Exponents have a well-known property:

$$x^ax^b = x^{a+b}$$

but

$$x^{a} + x^{b} \neq x^{a+b}$$

Similarly,

$$\log(a) + \log(b) = \log(ab) $$

But

$$\log(a)\log(b) \neq \log(ab)$$

So my question is this:

Is there a function $f$ on $\mathbb{R}$ or some infinite subset of $\mathbb{R}$ with the following properties

$$(1)\quad f(x)f(y) = f(x+y)$$
$$(2)\quad f(x)+f(y) = f(x+y)$$
ie
$$(3)\quad f(x)+f(y) = f(x)f(y)$$

It seems that $(2)$ requires the function to be linear...
Your title expresses interest in "a function in which addition and multiplication behave the same way". That's condition (3) alone. Conditions (1) and (2) are unnecessarily-strong requirements that artificially restrict the possible solutions. Be that as it may ...

Let's invoke condition (3) with three arbitrary values, $x$, $y$, $z$.

$$\begin{align}
f(x) + f(y) = f(x)\cdot f(y) \\
f(x) + f(z) = f(x)\cdot f(z)
\end{align}$$
Subtracting, we get
$$f(y) - f(z) = f(x)\cdot(\;f(y)-f(z)\;)\quad\to\quad\left(f(x)-1\right)\cdot\left(f(y)-f(z)\right) = 0$$
For all choices of $x$, $y$, $z$, at least one factor must vanish. We conclude that $f$ must be some constant; say, $k$. (The vanishing of the first factor requires specifically that $k=1$, but we'll go ahead and absorb this into the more-general statement.)

Then condition (3) reduces to 
$$k + k = k\cdot k \quad\to\quad k(k-2) = 0$$
so that $k = 0$ or $k = 2$. That is, we have two ways to satisfy condition (3):


  $$f(x) \equiv 0 \qquad\text{or}\qquad f(x) \equiv 2$$


Imposing conditions (1) and (2) limits the solutions to just the first.
Any functions where both sum and product = 2 will work. zero will not work because it results in division by zero. $(x+y)/(x*y)-1=0$