What is meant by "prove $M$ is a left $R$-module"? I'm sometimes asked to prove that certain abelian groups are $R$-modules for some ring $R$. Now to me this seems similar to being asked

Prove that the set $\{a,b,c\}$ is a group

Which doesn't make sense. Of course some operation can be defined on this set to make it into a group. However, by it self it just doesn't make sense to ask this question.
So my question is: Does there exist an abelian group $M$ and ring $R$ s.t. there does not exist a scalar multiplication that turns $M$ into an $R$-module? Since if this is the case, then I would understand being asked to prove that $M$ is an $R$-module, for this would not always be the case.

Here is an example of a question I would consider vague in this respect (although it's not the exact same problem as mentioned in the question, it's in the same spirit). 

$I$ a two sided ideal in $R$. Prove that $M$ is a left $(R\diagup I)$-module iff $M$ it is a left $R$-module that is annihilated by $I$.

To me this question doesn't make sense, because $M$ in itself is not an $R$-module; it's just an abilian group. So what does it mean to 'prove that $M$ it is a left $R$-module that is annihilated by $I$'. Does this means that there is SOME scalar multiplication that can be defined so that this holds? (in this case there is an obvious definition in terms of the multiplication of $M$ as an $(R\diagup I)$-module).  I still think that in this case the phrasing is misleading, because to me a module is really a triple: the group, the ring and the scalar multiplication. 
 A: First, I think you meant "...annihilated by $I$." not "....annihilated by $R$".
If someone says, "Show $X$ is a thing." and just specifies a set $X$ (without specifying any operations), there must be some understood obvious operations.
So if you are asked to show $\mathbb{R}$ is a group. It is understood that the group operation is $+$ ($\mathbb{R}$ has 2 obvious operations: addition and multiplication. Multiplication doesn't make $\mathbb{R}$ a group due to lack of inverses).
To answer your first question, yes. There are examples of pairs abelian groups and rings which do not allow for a module structure. For example: Let $R=\mathbb{R}$. Then any $\mathbb{R}$-module is a (real) vector space. So pick any finite abelian group, say $\mathbb{Z}_4$, and it cannot be a real vector space since real vector spaces are either trivial (sets of cardinality 1) or infinite (and $|\mathbb{Z}_4|=4 \not= 1$ or $\infty$).
For you next question, if $M$ is an $R$-module. Then there is a natural $R/I$ action iff $M$ is annihilated by $I$. Namely: $(r+I) \cdot m = r \cdot m$ for $r \in R$ and $m \in M$ (i.e. have cosets act via their representative).
This is the proper interpretation of your problem. The author wasn't explicit because "everyone knows" what the "correct" action "should" be. :)
To prove this: Assume $M$ is annihilated by $I$. Then verify that $(r+I) \cdot m = r \cdot m$ is a well-defined operation (i.e. equivalent representatives yield equal answers). Then run through the rest of the module actions to see that this actually makes $M$ an $(R/I)$-module.
For the converse: Assume $M$ is an $(R/I)$-module with this action. Then for all $x \in I$ and $m \in M$ you have $x \cdot m = (x+I) \cdot m = (0+I) \cdot m = 0 \cdot m = 0$ since $x+I=0+I$. Thus $M$ is annihilated by $I$.
EDIT: (To address your question from the comments)...
Let $M$ be an abelian group. If $M$ is also an $R$-module, the $R$-module structure doesn't have to be "internal" to the group. In fact, a single abelian group can have multiple module structures. 
For example: Consider $\mathbb{R}^2$ as an $\mathbb{R}[x]$-module. Let $\mathbb{R}$ act on $\mathbb{R}^2$ as scalar multiplication. Let $x$ act as your favorite linear endomorphism (linear operator) on $\mathbb{R}^2$. Each linear operator yields a different $\mathbb{R}[x]$-module structure (some may be isomorphic, some not, but all unequal) even though the underlying abelian group $\mathbb{R}^2$ is the same.
However, sometimes the module structure is determined by the group alone. This is true of $\mathbb{Z}$-modules. This happens because the $\mathbb{Z}$-action is forced to match additive-exponents. Consider for example: $3 \cdot x = (1+1+1) \cdot x = 1\cdot x +1\cdot x+1\cdot x=x+x+x=3x$. 
Each abelian group has exactly one $\mathbb{Z}$-module action. This is why texts identify $\mathbb{Z}$-modules and abelian groups as the "same thing".
