# It is possible to show/prove that the cancellation property is necessary to prove $0x=0$ for $x\not\in \mathbb{Z}^+ \cup \{0\}$?

For any nontrivial algebraic structures with additive identity 0 and multiplicative identity 1 (and binary operation defined by "juxaposition of its arguments"), and at least one sided distributive law holds, one can easily prove that 0 is an absorber, i.e. $0x=0$ for $x=\{0,1\}$ as follows:

$$0\cdot 1=0 \tag{Multiplicative identity}$$ and \begin{align} 0 &=0\cdot 1 & (\text{Multiplicative identity})\\ & =0\cdot (1+0) & (\text{Additive identity})\\ & =0\cdot 1+0 \cdot 0 & (\text{One sided distributivity})\\ & =0 + 0\cdot 0 & (\text{Multiplicative identity})\\ & = 0\cdot 0 & (\text{Additive identity})\\ \end{align} For $\mathbb{Z}^+$, one can prove $n\in\mathbb{Z}^+,0\cdot n=0$ by the following

\begin{align} 0 \cdot n&=0\cdot (1+1+1+\cdots ) & (\text{Inductive property})\\ & =0\cdot 1+0\cdot 1+0\cdot 1+\cdots & (\text{One sided distributivity})\\ & = 0+0+0+\cdots & (\text{Multiplicative identity})\\ & = 0 & (\text{Additive identity}) \end{align} But for $S=\mathbb{R},\mathbb{C}$ and general integral domains, given $x\in \text{S}$, the common proof is often seen as: \begin{align} 0 \cdot x&=(0 + 0)\cdot x & (\text{Additive identity})\\ & =0 \cdot x + 0 \cdot x & (\text{One sided distributivity})\\ & = 0 & (\color{red}{\text{Cancellation}}) \end{align}

Is the cancellation property necessary to prove $0$ absorbs $x\not\in \mathbb{Z}^+\cup \{0\}$. How can we sketch a prove to show this is a necessary condition?

It is not necessary. $0x = 0$ is true in any ring, not just a domain. That should indicate that cancellation not required.
To show $0x=0$, you need to show that $0x$ is an additive identity. Because addition is associative, the additive identity is unique, so it must be equal to 0. Thus, $$x + 0x = 1x + 0x = (1 + 0)x = 1x = x$$ demonstrates that $0x$ is an (i.e., the) additive identity.
More completely, consider that if $$a + b = a,$$ then $$a + b - a = a - a,$$ so $$b = 0$$ (note that I implicitly assumed associativity here). Set $a=x$ and $b=0x$ and the above shows that $0x = 0$.