Can we always define distributive multiplication If I have a set endowed with an addition operation (say a general group) and I know it contains 1, can I always define a multiplication operation so that it distributes over the addition? By this I don't require that the multiplication to even be associative but I wish for it to satisfy the condition that 1*a=a*1=a. Does it matter if addition was defined to be commutative?
 A: My idea is to steal a multiplication from an algebraic structure which already has distributive multiplication. More precisely, let our additive group be $(G,+)$. Assume that there exist a ring $(R,+,\cdot)$ with a nontrivial multiplication and an injective additive 
map $\phi: G \to R$. Then one can define $\phi^{-1}:\phi(G) \to G$ and this is again an additive map. If $\phi(G)$ is closed for the multiplication in $R$, then one can define a multiplication on $G$ by 
$$ a\cdot b:=\phi^{-1}(\phi(a)\phi(b))\quad (a, b\in G).$$
This will be a distributive multiplication.
If $R$ has an identity $1$ and $1\in \phi(G)$, then $\phi^{-1}(1)$ is the identity for the multiplication in $G$.
Of course, in the above reasoning, it is not necessary that we have a ring, it is enough that $(R,+)$ is an additive group with additional non-trivial binary operation which is distributive.
Of course, the above idea is not a solution of the posed problem. It is just a reformulation of the question.
One more comment, not related to the above thinking. If the multiplication in $(G,+)$ is
defined by $a\cdot b=a$, then we have distributivity from one side:
$$ (x+y)\cdot z=x+y=x\cdot z+y\cdot z\quad (x,y,z\in G). $$
A: For non-commutative multiplication, you cannot. Every element's order must divide the order of $1$: if the order of $1$ is $n$, then
$$
\underbrace{a+\dots+a}_{n\text{ times}} = a\cdot (\underbrace{1+\dots+1}_{n\text{ times}}) = a\cdot 0 =  0.
$$
(And $a\cdot 0$ has to be $0$ since $a\cdot 0 = a\cdot (0+0) = a\cdot 0 + a\cdot 0$.)
Now consider $S_3$: its elements have orders $1,2$ and $3$, so whatever element is chosen to be $1$, there would be elements such that their order does not divide its order.
