I do not understand this line of Wikipedia's page on the Basel problem:
If we formally multiply out this product and collect all the $x^2$ terms (we are allowed to do so because of Newton's identities), we see that the $x^2$ coefficient of $\sin(x)/x$ is...
Which Newton identity? How?
As Jerry says, it appears someone was trying to be pedantic.
What is meant is that for a polynomial $f(z)=z^n+a_{n-1}z^{n-1}+\dots+a_0$ with roots $\alpha_1, \dots, \alpha_n$, we have:
$$a_{n-1} = -\sum_{1 \leq i \leq n}\alpha_i$$ $$a_{n-2} = \sum_{1 \leq i_i < i_2 \leq n} \alpha_{i_1}\alpha_{i_2}$$ $$\dots$$ $$a_0 = (-1)^n\alpha_1 \dots \alpha_n.$$
You can see this by expanding $(z-\alpha_1)\dots (z-\alpha_n)$ and comparing with the coefficients of $f(z)$.
To solve the Basel problem, Euler did the same thing with the product expansion of $\frac{\sin z}{z}$, as if it were an infinite degree polynomial, and comparing with the Taylor expansion of $\frac{\sin z}{z}$.