Prove that $R$ is a ring with division. I'm having problems trying to know if my proof is wrong or not. The problem states:

Let $R$ a ring with 1, not necessary commutative, such that for every $a\in R\setminus\{0\}$, there exists $b\in R\setminus\{0\}$ (which depends on $a$) such that $a\cdot b=1$. Prove that $R$ is a division ring.

I have almost everything to prove that is a ring with division, I think I only miss the part that $b\cdot a=1$, what I have done so far is the following:
We have $ab=1$ so multiplying $b$ on the left side we have
$$b\cdot a\cdot b=b\cdot 1$$
Since $1$ is the $1$ of $R$, we get:
$$b\cdot a\cdot b=1\cdot b$$
(this is the part I'm not sure about) since we are in a group, we can cancel $b$ on the left side 
$$a\cdot b=1$$
and we are done?
 A: If the only part you are missing is to show that $ab = 1 \Rightarrow ba = 1$ then proceed as follows:
$$ab = 1 \Rightarrow bab = b $$
Now because of the statement, $b$ has a right inverse, $c $:
$$babc = bc \Rightarrow ba = 1$$
And we are done.
@Alex Wertheim also provides us with a different approach in the comment section. Be sure to check it.
A: All you need to show is that there also exists a left inverse, where for all nonzero $a$, there exists a $b$ so that $ba=1$. Well, you almost had it, the first step was fine, and the idea to "cancel" the $b$ was correct as well, you just need to use the hypothesis on $b$ as well:
let $a$ be nonzero.
Then there exists some $b$ so that $ab=1$. But then $bab=1 \cdot b$. Since $b$ is also nonzero, there exists some $c$ so that $bc=1$. Hence
$$babc=1 \cdot bc \implies ba=1.$$
A: I think the point is that $a$ and $b$ are non zero, also not zero-divisors. Hence to cancel $b$, you simply multiply $a$ from the left for both sides. However this problem is asking to prove $R$ is a division ring. To me, for every nonzero element in $R$ it has a multiplicative inverse has already been the definition. I don't see what to prove.
