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).