The answer to your question is essentially no: every non-degenerate associative binary operation which distributes over $+$ is multiplication in some (possibly non-commutative) rng. Let us abstract the situation a little: so suppose we have an abelian group $A$, whose operation we write as $+$. As with the addition of numbers, we have a zero element, commutativity, associativity, and negative elements.
Suppose we have a binary operation $\odot$ which distributes over $+$: in terms of equations, we have
\begin{align} a \odot (b + c) & = a \odot b + a \odot c \newline (a + b) \odot c & = a \odot c + b \odot c \end{align}
The first equation tells us that $a \odot -$ is a group homomorphism $A \to A$, and the second equation tells us that the map $a \mapsto (a \odot -)$ is a group homomorphism $A \to \textrm{End}(A)$. But $\textrm{End}(A)$ is naturally a (non-commutative unital) ring, and if we have the non-degeneracy condition
$a \odot b = 0 \text{ for all } b \text{ in } A \implies a = 0$
then every element $a$ is mapped to a unique endomorphism $a \odot -$ in $\textrm{End}(A)$. This allows us to identify $A$ with a subset A' of $\textrm{End}(A)$. The associativity condition
$(a \odot b) \odot c = a \odot (b \odot c)$
means we can identify $\odot$ with composition of endomorphisms, and this means that A' is a subrng of $\textrm{End}(A)$. If we demand that $\odot$ have a right unit, then the non-degeneracy condition is automatically satisfied; on the other hand, if we demand that $\odot$ have a left unit and be non-degenerate, then A' is even a subring of $\textrm{End}(A)$, and the left unit is also a right unit.