When addition distributes over multiplication Everyone is familiar with distributivity of multiplication over addition of real numbers. The distributivity of two binary operations sometimes goes both ways (e.g. max and min, or for lattices in general.)
Out of curiosity, I looked at the set of real numbers for which addition distributes over multiplication. A simple computation shows that this set is
$$\{a,b,c\in \mathbb{R}\, |\, a+b+c=1\}.$$
Is this just a meaningless fluke, or is there reason to expect something like this?
 A: For illustration purpose only, as this does not answer the question. However, I find this example quite instructive.
In a Boolean algebra, this short expression:
$$
b+ab'=a+b
$$
can be quite laborious to prove
\begin{align*}
b+ab'&=(a+a')b+ab' \\
            &=ab+a'b+ab' \\
            &=ab+ab+a'b+ab' \\
            &=a(b+b')+ab+a'b \\
            &=a+(a+a')b \\
            &=a+b
\end{align*}
However, if you do remember that in Boolean algebra + distributes over . the proof is immediate:
\begin{align*}
b+ab'&=(b+a)(b+b') \\
            &=a+b
\end{align*}
A: If you mean, "for which real numbers $a$, $b$, and $c$ do we have $a+(bc) = (a+b)(a+c)$?" then you get 
$$a+bc = a^2 + ab+ac + bc,$$ 
or $a=a(a+b+c)$. This means that either $a=0$, or else $a+b+c=1$, which is probably what you were trying to say (though you forgot the possibility that $a=0$).
But this is not a "set of real numbers for which addition distributes over multiplication". Such a set would be a collection of real numbers $X$, such that for all $r,s,t\in X$ you have $r+st = (r+s)(r+t)$, which would require that given any three elements in the set, either the first one you picked is $0$, or else they add up to $1$. In particular, any number you pick, picking it three times, would have to be either $0$ or $\frac{1}{3}$; but you cannot have both $\frac{1}{3}$ and $0$, because then picking $\frac{1}{3}$ for $r$ and $s$, and $0$ for $t$, you would not have $r+(st) = (r+s)(r+t)$. So the only collections $X$ that satisfy that condition are $X=\{0\}$ and $X=\{\frac{1}{3}\}$. Not very interesting at all...
So it's probably better to think about what you are looking for as the collection of all $3$-tuples of real numbers $(a,b,c)$ such that $a+bc = (a+b)(a+c)$, which consists exactly of all $3$-tuples with either $a=0$ or $a+b+c=1$. Geometrically you get the union of two planes in $3$-space: the $yz$-plane, and the $x+y+z=1$ plane. 
Nothing terribly exciting, I think, or particularly significant. But if you are interested in structures in which "addition" distributes over "multiplication" and vice-versa, then consider looking at boolean algebras, or more generally lattices, where $\wedge$ distributes over $\vee$ and vice-versa (like $\cap$ and $\cup$ do for sets, and the logical operators AND and OR do for logic). 
A: This is a reasonable question to ask. Like "What is the factorial of a non-integer?" or "Can you take the derivative of a regular expression?". Those turn out to have extensions to other nearby domains (in opposite directions).
But with axioms the mind bending is of a different sort. Axioms, like distributivity, capture what we know about a domain. The domain actually comes first and we discover what the axioms are that preserve truths in that domain. If you change an axiom you're actually getting a new domain. Things mean something different. If you change the 5th axiom (the parallel postulate) of the 5 axioms of Euclidean geometry, you get a different geometry. The new lines aren't exactly  what the traditional Euclidean lines were.
Distributivity of multiplication over addition is true -in real numbers-. If everything is kept the same, those numbers for which addition distributes over multiplication  turn out to be a subset of the reals (as others noted, if $a+(b*c) = (a+b)*(a+c)$ then either $a = 0$ or $a+b+c = 1$).
But if you want distributivity to happen in both directions, -and- always, then you can think of it in two ways:


*

*It's now just not the case (because maybe you wanted to stay in the reals and the additional distributivity rule doesn't work on all of the reals). That is, reverse distributivity isn't universal for the reals.

*Or now you're working with something that is -not- the reals. The models that satisfy both axioms are just not like the reals at all (actually it's true of the traditional + and * and 0 and 1, and also true of other things called boolean algebras. You now have a different system you're working with.
So if you change the axioms you're getting a different subset of what you originally had, or a better way to think of it is that you're now in a (maybe slightly, maybe a lot) different universe where the rule may just not work like in the old one.
So adding an axiom doesn't produce a meaningless fluke. Well, it may have (the axiom may have accidentally forced a contradiction and then have no models), but in this case it didn't. Usually it just produces a system that is ... different, maybe similar but just not the same thing.
A: Addition may not be generally distributive over multiplication, but operators that are do exist.  For example the operator # defined by
a#b = exp(ln(a) * ln(b))
is distributive over multiplication:
a#(b*c) = (a#b) * (a#c)
(BTW, # so defined is also associative & commutative, and has a well-defined identity element and # inverses.  * and # can be manipulated to form a pretty variation on traditional calculus.)
Turning it around, the operator % defined by
a%b = ln(exp(a) + exp(b))
is distributed over by addition:
a+(b%c) = (a+b) % (a+c)
