If $ab = e$, then $ba = e$? $x\cdot (a\cdot b) = x\cdot e$
$(x\cdot a)\cdot b = x$
$e\cdot b = x$
$b = x$
Are my steps correct?
What I wanted to prove is that if $ab = e$, then $ba = e$
$x$ is inverse of $a$ and $e$ is identity element .
 A: It depends on the structure from where you took the elements. If it's a group, then it's fine. For an algebraic structure where right identity $\neq$ left identity, or right inverse $\neq$ left inverse, your proposition won't be true.
A: I don't see why you need to introduce $x$ for $a^{-1}$. Since $ab=e$ we have
\begin{align*}
e & = a^{-1}ea\\
 & = a^{-1}(ab)a\\
 & = ba
\end{align*}
by multiplying 
by $a^{-1}$ from the left and then by $a$ from the right. So conjugating $ab=e$ gives $ba=e$.
A: I'm not that sure what you did was correct. As always with maths, it depends a lot on the context.
If your context is that $a, b, x$ are elements of a group $G$, than it still depends how you did define group. To me it appears like you want to prove that if every element $a \in G$ has a right-inverse $b\in G$, i.e. $ab = e$, then $b$ is also a left-inverse.
If this is what you want to do, then I don't see why your second step is true. You seem to use $x\cdot a = e$, but why is this true?
On the other hand, if you suppose the existance of a right-inverse $b$ and a left-inverse $x$ than it looks fine.
A: One proof if you are referring elements in a group, $a,b,e\in G$ and $ab = e$. Then
$$ba = bea = baba$$
The group $G$ is closed, thus $ba\in G$. Every element in $G$ has an inverse, thus $(ba)^{-1} \in G$. Then by associative property
$$e = (ba)^{-1} ba = [(ba)^{-1} ba ]ba = ba$$
A: In a group $G$ one has the general property that for all $a$, $b\in G$ the elements $ab$ and $ba$ are conjugated. Indeed one has
$$
ba=b(ab)b^{-1}.
$$
What is asked follows immediately from the observation that the only element conjugated to the identity $e$ is $e$ itself.
