Let $S$ be a nonempty set, and let $R_1=(S,+,\cdot)$ and $R_2=(S,+,\star)$ be two rings. Suppose that $R_1$ and $R_2$ have the same multiplicative identity element $1$, and that $1$ generates the group $(S,+)$. Does it follow that $R_1=R_2$?
My question is motivated by considering how powerful the distributive property is in determining the multiplication table of a ring. If $\alpha,\beta\in R_1$ and $\alpha=\underbrace{1+\dots+1}_{\text{$m$ times}}$ and $\beta=\underbrace{1+\dots+1}_{\text{$n$ times}}$, then \begin{align} \alpha\cdot \beta &= \left(\underbrace{1+\dots+1}_{\text{$m$ times}}\right)\cdot\left(\underbrace{1+\dots+1}_{\text{$n$ times}}\right) \\[5pt] &= \underbrace{(1\cdot 1)+\dots+(1+1)}_{\text{$mn$ times}} \\[5pt] &= \underbrace{1+\dots+1}_{\text{$mn$ times}} \end{align} Therefore, it seems that if two elements can be obtained by repeatedly adding $1$ to itself, then their product is determined by distributivity. However, I am not sure about the case where an element $x\in S$ is generated by repeatedly adding $-1$ to itself.
Your approach works exactly the same way if one or both of the elements are sums of $-1$ instead of sums of $1$. For instance, if $\alpha$ is the $m$-fold sum of $-1$ and $\beta$ is the $n$-fold sum of $1$, then by distributivity their product is the $mn$-fold sum of $(-1)(1)=-1$, and thus fixed.