Assume that $R$ is a ring that might not have a unit element. What is the correct answer to the following question:
If $N\subset M$, where $M$ is an $R$-module and $rn_1 + n_2 \in N$ for all $n_1,n_2 \in N$ and $r\in R$, then is $N$ a submodule?
Recall that a subset $N$ of a module $M$ is called a submodule if it is an additive subgroup and $rn$ in $N$ for all $n \in N$ and $r \in R$.
The condition $rn$ in $N$ for all $n \in N$ and $r \in R$ is easy to verify, but it is not clear if $N$ is a subgroup.
According to Topics in Algebra by Herstein, modules can also be considered over rings that do not have a unit element.
If one does not impose that the module is unital that's not true anymore. (Whether the ring has a unity or not is not really crucial.)
Let $M=\mathbb{Z}^2$ with the usual addition yet scalar multiplication by elements from $\mathbb{Z}$ is defined as follows: $r (x_1, x_2) =(rx_1,0)$.
This verifies all conditions in the definition you recall.
Now, the set $\{(x,y) \colon x \in R, \, y \in \{0,a\}\} \subset M$, where $a$ is some non-zero integer, is a set that verifies the condition in the question yet it is not an additive subgroup.
The very same construction works for other rings too. That the ring has an element whose additive order is greater than $2$ is also that's needed. Thus, we could also work with the ring of even integers for example.