I hit a slight snag when trying to prove that the inverse function $x\mapsto x^{-1}$ on the unit group is continuous in a ring with an absolute value, so I'd like some confirmation that the theorem is in fact true.
Here's a version of the classical proof that reciprocal is continuous on $\Bbb C\setminus\{0\}$:
Let $x\ne 0$ and $\epsilon>0$, and let $\delta=\dfrac{|x|}2\min(1,\epsilon|x|)$. Now suppose $y\ne0$ and $|x-y|<\delta$. Then $$|y|\ge|x|-|x-y|>|x|-\delta\ge\frac{|x|}2,$$ so $|x-y|<\delta\le\dfrac{|x|}2\epsilon|x|<\epsilon|xy|$. Thus $\epsilon>\dfrac{|x-y|}{|xy|}=\left|\dfrac {x-y}{xy}\right|=\left|\dfrac1x-\dfrac1y\right|$.
Most of this proof is interpreted easily in an arbitrary ring with an absolute value (replacing "$x\ne0$" with "$x$ is a unit"), but I can't complete the final step. We have $|xy||x^{-1}-y^{-1}|=|xyx^{-1}-xyy^{-1}|=|xyx^{-1}-x|$, but there seems no reason to believe that this is equal to $|y-x|$ unless I assume the ring is commutative.
Is it true that the reciprocal function in a normed ring is continuous? What if we assume that the ring is a division ring?
Edit: By "normed ring" or "ring with an absolute value" I am referring to Absolute value (algebra), which is to say a map $|\cdot|:R\to\Bbb R$ satisfying $|x|=0\iff x=0$, $|x-y|\le|x|+|y|$, $|xy|=|x||y|$.
The key here is that the real numbers form a commutative ring. Therefore $$ |xy||x^{-1} - y^{-1}| = |x||y||x^{-1} - y^{-1}| = |x| |x^{-1} - y^{-1}||y| = |xx^{-1}y - xy^{-1}y| = |y-x|. $$ (P.S. I assumed that by normed ring you meant Banach algebra, which puts an inequality ; I only put equality because you did. Was I correct? If the answer is not helpful please take the time to add a definition.)
Hope that helps,