About rings with a right multiplicative identity $1$ such that every element in $R\setminus\{0\}$ has a left-inverse 
$R$ be a  ring with a right multiplicative identity $1$ such that for every $a \in R \setminus \{0\}$, $\exists x \in R$ such that $xa=1$ i.e. every element in $R \setminus \{0\}$ has a left-inverse. Then is it true that $R$ is a division ring ? 

(I know that left-identity and left-inverse implies group similarly right-identity and right inverse implies group, but am not sure about the other combination.) 
I have tried for every  $a,b \in R$ \ $\{0\}$ $a=a.1=a(x_b.b)=(a.x_b).b$ ; if for some $b$ , $a.x_b=0 $ then $a=0.b=0 \in R$ \ $\{0\}$ impossible ; so $a=yb$ has a solution for $y \in R$ \ $ \{0\}$ for every $a,b \in R$ ; but I cannot show solution for $a=b.x$ . Please help . 
 A: (We must add to the assumptions that $1\ne0$, or the assertion would be false, because the zero ring is not a division ring.)
Let's first show that a right multiplicative identity is also a left identity.
Call $\rho$ a right multiplicative identity: $a\rho=a$, for every $a\in R$. If $\rho a-a\ne0$, there exists $x$ such that $x(\rho a-a)=\rho$. Now
$$
x(\rho a-a)=(x\rho)a-xa=xa-xa=0,
$$
a contradiction. So $\rho a=a$. Hence $\rho$ is also a left identity.
Now we can set $\rho=1$ and use it as a two-sided identity.
First of all, if $xa=1$, then $x\ne0$. Since $x\ne0$, it has a left inverse, call it $y$: $yx=1$.
Thus
$$
y=y1=y(xa)=(yx)a=1a=a
$$
and so $x$ is both a left and right inverse of $a$.
A: First of all $1 \ne 0$, because if $1=0$, then $a=a\cdot 1=a\cdot 0=0 , \forall a \in R$ and $R\setminus\{0\}$ would be empty, not possible. 
Now it is given for every $a\in R\setminus\{0\}$, $\exists a' \in R$ such that $a'a=1$. 
If for some $x\in R\setminus\{0\}$, $x'=0$ then $1=x'x=0\cdot x=0$, impossible. 
So for every $a \in R\setminus\{0\}$, $\exists a' \in R\setminus\{0\}$ such that $a'a=1$, so also $\exists a'' \in R\setminus\{0\}$ such that $a''a'=1$, hence 
$a'a''=a'(a''\cdot1)=a'\big(a''(a'a)\big)=a'\big((a''a')a \big)=a'(1\cdot a)=(a'\cdot 1)a=a'a , \forall a \in R\setminus\{0\} $, so $a'(a''-a)=0$, whence 
$1\cdot(a''-a)=(a''a')(a''-a)=a''\big(a'(a''-a)\big)=a''\cdot0=0 , \forall a\in R\setminus\{0\} $. 
Now if $a''-a \ne 0$, then $\exists (a''-a)' \in R\setminus\{0\}$ such that $(a''-a)'(a''-a)=1$, whence 
$1=(a''-a)'(a''-a)=\Big((a''-a)'\cdot1\Big)(a''-a)=(a''-a)'\Big(1\cdot(a''-a)\Big)=(a''-a)'\cdot0$ $=0$, impossible! 
So $a''-a=0$ i.e. $a''=a, \forall a \in R\setminus\{0\} $; thus for every $a \in R\setminus\{0\}$, $\exists a' \in R$ \ $\{0\}$ such that $a'a=1=aa'$. 
Now $0\cdot 1=1\cdot 0=0$ and to show that $1 \in R\setminus\{0\} $ is an identity of $R\setminus\{0\}$ we note that for every $a \in R\setminus\{0\}$, $1\cdot a=(aa')a=a(a'a)=a\cdot1=a$. 
Lastly to show that $R\setminus\{0\}$ is closed under $\cdot$ we note that $a\ne0$ and $ab=0 \implies b=1\cdot b=(a'a)b =a'(ab)=a'\cdot0=0$; $b \ne 0$ and $ab=0 \implies a=a\cdot 1=a(bb')=(ab)b'=0\cdot b'=0  $, thus $a\ne0 , b\ne0 \implies  ab \ne 0$, hence $( R\setminus\{0\}, \cdot)$ is a group.
