If your group has the property that every element has finite order, but there is no upper bound on the orders of the elements, then it is not the additive abelian group of a ring with identity. The reason is that if there were such a ring structure with an identity $1$, then $1$ would have finite additive order $k$, and then for all $a$ in your group, $k\cdot a=(k\cdot1)a=0a=0$, which forces $a$ to have order at most $k$.
For each prime $p$, the Prüfer $p$-group $\mathbb Z(p^\infty)$ is an example of such a group. The quotient group $\mathbb Q/\mathbb Z$ is another. Direct sums (but not direct products) of infinitely many finite cyclic groups of unbounded order would also be examples.
Let $(G,\circ)$ be a group with neutral element $e$ and let $X$ be a set. We say that $G$ acts on the set $X$ if we are given a map $G\times X\to X$, $(g,x)\mapsto g\cdot x$ if the following natural requirements are fulfilled: $(g\circ h)\cdot x = g\cdot(h\cdot x)$ and $e\cdot x=x$ for all $g,h\in G$, $x\in X$. (The first requirement may allow one to be sloppy and use the same multiplication symbol for the group operation and the action and also drop parentheses).
If $X$ carries additional structure, say $(X,*)$ is also a group, we - naturally - demand that more requirements be respected before we say that the group $(G,\circ)$ acts on the group $(X,*)$, namely that $g\cdot(x*y)=(g\cdot x)*(g\cdot y)$. Note that this looks like the distributive law, but $g,x,y$ are (usually) not from the same set!
If $(X,*)$ is an abelian group with neutral element $e$, then there is always a standard way to define an action of the additive group $(\mathbb Z,+)$ on it:
Define $0\cdot x=e$ first, then by the recursion $(n+1)\cdot x=(n\cdot x)*x$ define the action of all positive integers and finally by $(-n)\cdot x=(n\cdot x)^{-1}$ the action of negative integers. (Verify that this is an action, be careful with all those different operation symbols!).
If we are a bit sloppy and use $+$ also for the group operation of $X$ (and negation for inverse), as is usual for abelian groups, then the fact that $(\mathbb Z,+)$ acts on the group $(X,+)$ precisely formalizes that "distributivity" holds.
As $(\mathbb Z,+)$ is also an abelian group, the above method can be used to define an action of $(\mathbb Z,+)$ on itself. We should be careful because we use $\cdot$ to denote this action and already have an intrinsic multiplication of integers that is also written with $\cdot$. Fortunately, the group action of $(\mathbb Z ,+)$ on itself is precisely the normal integer multiplication (check that without getting confused about the different operations).
As a bonus check that $(n\cdot m)\cdot x=n\cdot (m\cdot x)$ holds for all $n,m\in \mathbb Z$ and $x\in X$ with the group action of $(\mathbb Z,+)$ on the abelian group $(X,*)$ defined above, justifying again the use of the same symbol for both things (and allowing one to drop parentheses) even though they are strictly speaking not the same.
Best Answer
Here is a "dumb" example. Let $R=\mathbb Z$, and let $\times=+$, i.e., addition and multiplication are the same thing. Now $(R,\times)$ is a commutative monoid, with a $1$ (i.e, $0\in R$). This is clearly not distributive: $1\times(1+1)=3\neq1\times 1+1\times 1=4$.