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$A$ is a ring, $a,b\in A$. How to show that $ab$ and $ba$ are invertible $\Rightarrow a$ and $b$ are invertible?

This is how I did it:

Let $x$ be the inverse of $ab$ and let $y$ be the inverse of $ba$. Therefore $x(ab)=(xa)b=I$ and $(ba)y=b(ay)=I$. And now $(xa)b=I=b(ay)$ implies that $b$ is invertible. Analogically I can show that $a$ is invertible.

Is this correct?

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    Completely correct :) – 2017-02-17
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    @user394691 You can prove that $ay=xa$ the same way you prove this lemma: "Let $(X,\cdot)$ be a monoid. Let $x$ be an element such that there are $y,z\in X$ who satisfy $yx=1=xz$. Then, $y=z$. Moreover, $y$ is the only element in $X$ which satisfies either $yx=1$ or $xy=1$." – 2017-02-17
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    @user394691 Technically you need to prove it. Whether you have to actually write the proof or not, it depends on the context in which the question arose: it is an easy, well known fact that an element in a monoid is invertible if and only if it is both invertible on the left and on the right, so one is allowed to invoke it (for instance, a proof in a book might go back to it). However, the person who gave you the assignment might want to see you do the verification. – 2017-02-17

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To complete your proof (that $b$ is invertible), all you need to argue is that, since $xab=bay=1$, then

$$xa=(xa)(bay)=(xab)ay=ay$$

(with a similar argument for $a$).