Transformation rules of a linear operator $S\mathbf y = A \mathbf x \implies \mathbf y' = AS \mathbf x'$

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Given are the equations

$$ \mathbf x = S \mathbf x' \tag{8.91} $$ $$ \mathbf y = A \mathbf x, \quad \mathbf y' = A'\mathbf x' \tag{8.93} $$

It (Riley Hobson Bence, 3rd) then says,

But using $(8.91)$, we may rewrite the first equation as $$ S\mathbf y' = AS\mathbf x' \implies \mathbf y' = S^{-1} AS \mathbf x'. $$

Now, intuitively, I believe that $\mathbf y' = S^{-1} AS \mathbf x'$ holds. Also, I don't dispute the implication above; it just results by multiplying $S^{-1}$ on the left of the left hand side.

But I'm at odds on how the text arrives at $S\mathbf y' = AS\mathbf x'$, that is, without assuming that $\mathbf y' = S^{-1} AS \mathbf x'$ holds. How does this follow from $(8.91)$, like the text says?

In other words, I'm unable to derive how we arrive at $S\mathbf y' = AS\mathbf x'$, without assuming $\mathbf y' = S^{-1} AS \mathbf x'$.

I've tried the following:

$$ \mathbf y'= A'\mathbf x' \tag{1} $$ Multiplying by $S$, we get $$ S \mathbf y' = SA'\mathbf x' \tag{2} $$ $(2)$, is in disagreement with the text $S\mathbf y' = A'S\mathbf x'$

Edit. Removed the second part of the question, which included a typo.

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A missing piece of information, it seems, is that (8.91) is also supposed to tell us something about $\mathbf y$. In particular, we have $$ \mathbf y = S\mathbf y'. $$ With that, we have $$ \mathbf y = A \mathbf x \implies S\mathbf y' = A (S \mathbf x') \implies \mathbf y' = S^{-1}AS \mathbf x'. $$