Question about product of generating functions in a proof that for positive integer $n$, $\sum_{k=0}^n(-1)^k\binom nk\binom{2n-k}n=1$.

I’d carry out the last step a bit differently. What the first part shows is that $S_n$ is the coefficient of $x^n$ in the product

$$(1-x)^n\cdot\frac1{(1-x)^{n+1}}\;,\tag{1}$$

something that is often written

$$S_n=[x^n]\left((1-x)^n\cdot\frac1{(1-x)^{n+1}}\right)$$

with the $[x^n]$ operator. Clearly, then,

$$\begin{align*} S_n&=[x^n]\left(\frac{(1-x)^n}{(1-x)^{n+1}}\right)\\ &=[x^n]\left(\frac1{1-x}\right)\\ &=[x^n]\sum_{k\ge 0}x^k\\ &=1\;. \end{align*}$$

The $n$ in $(1)$ really does depend on which $S_n$ we’re computing, but $(1)$ simplifies to $\frac1{1-x}$ for all $n$, so in the end we really are looking at one power series.