Can every set be a group? Lets consider a set A. Can we prove that  there exists at least one operation * such that under it the set forms a group. May I know how to approach this question?
 A: The answers are already given, but since you asked how you should approach this, I'll try to address that issue. 
The first thing to understand is that if you have a bijection $f:A\to B$ between any two sets, then any group structure on $A$ can be transferred along $f$ to a group structure on $B$. More precisely, there is precisely one group structure on $B$ for which $f$ is a group homomorphism (necessarily isomorphism). If this is not clear to you, spend some time and prove it. 
Now, the question becomes equivalent to, given any cardinality $\kappa$ (other than $\kappa = 0$), is there some set of that cardinality which supports a group structure. (Under the axiom of choice,) every set has a cardinality and so if for every cardinality there is some  group of that cardinality then, using the result above, every set supports a group structure (upon choosing a bijection, so the axiom of choice is used again). 
So, it boils down to constructing groups of all possible cardinalities. Finite ones are dealt with easily by the finite cyclic groups. Infinite ones are constructed by taking products of sufficiently many finite groups, using basic rules of cardinal arithmetic (assuming the axiom of choice again).  
A: See this: https://mathoverflow.net/questions/12973/does-every-non-empty-set-admit-a-group-structure-in-zf
It is equivalent to the axiom of choice that every nonempty set has a group structure.
A: I can do you one better. Every set has a (somewhat natural) ring structure! You surely know what to do for finite sets. Suppose now that $X$ is an infinite set, and let $S$ denote the set of all finite subsets of $X$. Then, it's easy to prove that $S$ is a ring with addition being symmetric difference, and multiplication being intersection. Since $\#(X)=\#(S)$ you can find a bijection $f:S\to X$ and so pushforward the ring structure to $X$.
A: It is not clear to me why precisely you would want to do this, but the answer is YES. Assuming Axiom of Choice (which is done in most of mainstream mathematics, but may not be done in the more foundational parts) your question is equivalent to asking if there is a group of the same cardinality as $A$.
If $A$ is finite, with cardinality say $m$, then you have a finite group with the same cardinality: namely the cyclic group of order $m$, which is sometimes written $C_m$ or $\mathbb{Z}/m\mathbb{Z}$.
If $A$ is infinite with cardinality $\alpha$, then you can just use the fact that infinite cardinalities are rather "persistent" under various operations. If $G$ is your favourite finite group, then the group $\bigoplus_{\iota < \alpha} G$ (direct sum of $\alpha$ copies of $G$) has the same cardinality as $A$. (Such direct sum consists of sequences $(x_\iota)_{\iota < \alpha}$ with $x_\iota = e_G$ for all but finitely many $\iota$. The operation group operation is just the pointwise operation)

To avoid being outdone by Alex, let me mention that $\mathbb{Z}/m\mathbb{Z}$ is actually a ring. Also, a direct sum of $\alpha$ copies of a given finite ring is a ring with cardinality $\alpha$, for $\alpha \geq \aleph_0$. So the above can be adapted to show that $A$ can be given a ring structure. Of course, Alex's argument is more elegant for doing this.
