Proof with fields. I'm stuck with a problem on my Algebra study.

Let $R$ be a commutative ring with more than one element. Prove that
  if for every nonzero element $a \in R$, we have $aR=R$, then $R$ is a
  field.

I've been brainstorming for a while. So, this is what I have so far:
Since $aR=R$  $\forall a \neq 0 \in R,r \in R$, then $ar \in  R$ (obvious step).
But since the subset of a non-zero is contained in $R$, then all elements $a$ are in $R$. (I think this might be important).
If $R$ had the inverse element $r^{-1}$ $ \forall r$, it would be trivial, but it's not a ring requirement to have a multiplicative inverse (unit). So I'm stuck here. I don't think I can guarantee that ar will ever produce the identity element.
Thanks in advance!
 A: As pointed out in the comments I ignored some of the finer details of the question in my original answer, so here is a fixed version:

As dREaM pointed out in the comments every such ring has a unit: For a fixed non-zero element $a \in R$ we have $a R = R$, so there exists some element $1 \in R$ with $a \cdot 1 = a$ (because $R$ contains more than one element such a non-zero element does exists). For an arbitrary element $b \in R = aR = Ra$ there exists some $r \in R$ with $b = ra$, which is why
$$
 b \cdot 1 = ra \cdot 1 = ra = b.
$$
So $1$ is the claimed unit. Notice that because $R$ contains more then one element we have $1 \neq 0$.
For every non-zero $a \in R$ we have $aR = R$, so in particular there exists some $b \in R$ with $ab = 1$. This shows every non-zero element in $R$ is invertible. So $R$ is a unitial, commutative ring with $1 \neq 0$, such that every element $a \in R$ is invertible, i.e. $R$ is a field.
A: Hint: If $aR = R$, then in particular there is some $r$ so that $ar = 1$.
Exercise (developing this idea): If $R$ is a domain, and also a contains a field $k$ as a subring, so that $R$ is a finite dimensional vector space over that field (the field $k$ acts on $R$ by multiplication), then $R$ is a field.
