Fields work that way, more generally domains. The additive order of every nonzero element of a domain $R$ is the same.
Proof: Let $n$ be additive order of $1_R$, i.e. the smallest positive integer such that $n \cdot 1_R = 0_R$ (note that every ring is a $\mathbb Z$-algebra so we can multiply with integers). Then every element $x\in R$ with $x \neq 0_R$ satisfies $n \cdot x = n \cdot 1_R \cdot x = 0_R \cdot x = 0_R$, so the additive order $k$ of $x$ divides $n$. But if $k < n$, then $0_R = k \cdot x = (k \cdot 1_R) \cdot x$, so $x$ is a right zero divisor. But domains do not have zero divisors, so $k = n$.
The previous argument assumed that $1_R$ has finite order. But if the order of $1_R$ is infinite, then a similar argument shows that the order of all nonzero $x$ must be infinite.
But if $R$ is not a domain, this is not true in general. You've found an example yourself.