Alternate translation for: “Every real number except zero has a multiplicative inverse.”

Yes, it is a general principle that if $y$ does not appear in $\varphi$, then the following are equivalent.

  1. $\varphi \rightarrow \exists y(\psi)$
  2. $\exists y(\varphi \rightarrow \psi)$

Yes :

$∀x((x \ne 0 ) → ∃y(xy = 1 ))$

and

$∀x∃y((x \ne 0) → (xy = 1))$

are logically equivalent, because :

$\vdash \exists y (\alpha \rightarrow \beta) \leftrightarrow (\alpha \rightarrow \exists y \beta) \quad $ if $y$ is not free in $\alpha$.

In your case, $\alpha$ is $(x \ne 0 )$ and $y$ is not free in it.