Doubt on proofs of "If $A, B$ are $n\times n$ matrices such that $AB = I$ then $BA = I$" The proofs are already presented at length here. And I found that most of the solutions used the fact that if the linear map $B$ is one-one then it must be onto.
But this fact appears trivial because of $AB = I$. Consider the equation $Bx = y$ and if there are two solutions $x, x'$ then $Bx = Bx' = y$ by multiplying by $A$ on left we get $x = x'$. This is what is referred to as $B$ is one-one. But the same argument also shows that $x = x' = Ay$ so that $B$ is onto (for each $y$ we have found $x = Ay$). I wonder why do the proofs in the linked question used some linear-algebra to show that $B$ is onto.

It appears that most proofs use the left multiplication by $A$ to ensure that $x = x'$ but somehow this does not seem to guarantee $x = Ay$. Why is that the case? Or am I missing something?

Update: Thanks to all those who replied. I got hold of my mistake. The argument proves that if $Bx = y$ has a solution it must be $x = Ay$. It does not show that there is a solution. Sorry to bother you all for such trivial matter (and if all agree I may delete this silly question). Existence of a left inverse of $B$ does not guarantee the solution of $Bx  = y$. Rather it is the existence of right inverse of $B$ which is needed here to get a solution.
 A: 
But the same argument also shows that $x = x' = Ay$ so that $B$ is onto (for each $y$ we have found $x = Ay$)

It's not totally clear to me what you mean by that sentence.  Note, however, that in the beginning of that first argument, we say if there are two solutions $Bx = Bx' = y$.  We cannot assume (without more information) that there any solutions to $Bx = y$.
Note that there are certainly matrices that have a left inverse, but fail to have a right inverse.  For instance, with
$$
B = \pmatrix{1&0\\0&1\\0&0}
$$
Then $A = B^T$ is a left inverse, so that $B$ is one to one.  It is clear, however, that the transformation $x \mapsto Bx$ fails to be onto.  Clearly, your argument must go wrong somewhere.
It is also notable that if $V$ is infinite dimensional, then a linear $T:V \to V$ may be one to one but not onto.  As a classic example, consider the map $f(x) \mapsto \int_0^x f(t)dt$.
(I like Jonas's answer a lot.  Looking at arbitrary functions allows for a useful analogy).
A: (There are some parallels with Omnomnomnom's answer, and we even chose the same quote to start our answers.  This note helps visually distinguish them.)

But the same argument also shows that $x = x' = Ay$ so that $B$ is onto (for each $y$ we have found $x = Ay$).

To show $B$ is onto, you need to show that for each $y$ there exists $x$ with $Bx=y$.  Your claim is that $x=Ay$ will do, so let's see if this works: $Bx=BAy=\cdots$?  We'd like to say that's $y$, but it assumes the result that $BA=I$.
You have to use linear algebra because for functions $f,g:X\to X$, $f\circ g(x)=x$ for all $x\in X$ implies that $g$ is one-to-one and $f$ is onto, but does not imply that $g$ is onto or that $f$ is one-to-one.  For example, $X=\mathbb R$, $g(x)=e^x$, $f(x)=\ln(x)$ when $x>0$, $f(x)=0$ when $x\leq 0$.
There are linear counterexamples on infinite dimensional spaces. Let $V$ be the space of real sequences $(a_1,a_2,\ldots)$, and let $A$ and $B$ be the left and right shifts, $A(a_1,a_2,a_3,\ldots)=(a_2,a_3,a_4,\ldots)$, $B(a_1,a_2,a_3,\ldots)=(0,a_1,a_2,\ldots)$.  Then $AB=I$, but $BA\neq I$.
