In a ring $R$ and $I$ is an ideal of the ring, we have a result that if $I$ contains a unit then the ideal is equal to the whole ring and since the moment the ideal contains a unit it won't be just an ideal which means there does not exist an ideal with only an unit?
I am not able to express the question with 100% accuracy, still can someone help?
When talking about ideals we have two 'trivial' cases: first, the zero ideal $(0)$, and second, the whole ring $R$ viewed as ideal. Both play a special role (e.g. a ring is a field iff its only ideals are these two) but are somewhat not what you usually think of as ideals. But, there is no restriction placed on an ideal $I$ being proper per se, that is being a proper subset and not just the whole.
In fact, there result you stated is slightly stronger
Claim. Let $R$ be a ring. $a$ is a unit in $R$ if and only if $(a)=R$.
Proof. If $a$ is a unit, there is some $b\in R$ such that $ba=1$, thus $1\in(a)$. But then for all $r\in R$ we can write $r=r.1=r(ba)=(rb)a\in(a)$ and therefore $(a)=R$. Conversely, assume $(a)=R$. Then in particular $1\in(a)$ and there is some $b\in R$ such that $ba=1$. That is nothing else than $a$ being a unit and we are done. $\square$
So, yes, there is no proper ideal containing a unit as every such ideal is already the whole ring. That is why when talking about, let's say, maximal ideals we explictely define them as being proper. Otherwise we would have a trivial maximal ideal for every unit and that would screw up some things.