Wherever it converges, the Riemann zeta function $\zeta(s)$ is equal to the $p$-series - and it converges only when the real part of $s$ is greater than one. The Euler product analogously only converges when $\mathrm{Re}(s)>1$, so the reasoning that $\zeta=1/\text{something}\ne0$ does not apply where the product is an invalid representation of zeta.
Moreover, the idea is flawed all by itself: the "$\text{something}$" is an infinite product, so if that product diverges to infinity in the limit, we have $\text{something}^{-1}\to0$. For example, formally we have
$$0\le\frac{1}{(1+1)(1+1/2)(1+1/3)\cdots}\le\frac{1}{1+1/2+1/3+\cdots}=\frac{1}{\infty}=0.$$
In general, properties (e.g. inequalities) satisfied by expressions with a finite number of terms do not necessarily carry over to an infinite number of terms (but we can still apply logic to the partial sums if need be).
So if neither the original series nor the product converges outside the right side of $1$, how can it be then that $\zeta(s)$ can be computed on or to the left of this? The answer is analytic continuation, which I gave a brief introductory explanation to just recently. But then precisely how can it be explicitly extended left of $1$? The simplest, but limited, way is through the Dirichlet eta function:
$$\eta(s)=1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}-\cdots=\left(1+\frac{1}{2^s}+\frac{1}{3^s}+\cdots\right)-2\left(\frac{1}{2^s}+\frac{1}{4^s}+\cdots\right)$$
$$=\zeta(s)-\frac{2}{2^s}\zeta(s)=(1-2^{1-s})\zeta(s).$$
The eta function converges for $\mathrm{Re}(s)>0$, so if we write $\zeta(s)=(1-2^{1-s})^{-1}\eta(s)$ we have a way to evaluate zeta to the left of where we originally could (by $1$ at most). Analytic continuation beyond this requires more sophisticated machinery - a functional equation was first established by Bernhard Riemann in the paper "On the Number of Primes Less Than a Given Magnitude,"
$$\zeta(s)=2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\zeta(1-s).$$
This means that what $\zeta$ evaluates to on the left of $s=1/2$ is determined by its evaluation on the same point reflected across $s=1/2$. With this and the $\eta$ continuation we are given the ability to compute $\zeta$ anywhere we like. Note that with the presence of the $\sin$ above, it is trivially true that $s=-2n$ are zeros of $\zeta$ for $n=1,2,3,\dots$ Moreover, since $\zeta(s)$ has no zeros right of $\mathrm{Re}(s)=1$, the functional equation predicts it has no other nontrivial zeros to the left of $\mathrm{Re}(s)=0$: this means all of the nontrivial zeros lie in the critical strip.
The Riemann Hypothesis is that all nontrivial zeros have real part $1/2$, but it has not been proven - we can, however, calculate the zeros to arbitrary accuracy and prove when particular zeros have real part exactly half, see this question linked by J.M.
As a sidenote, while the Euler product doesn't converge globally, I believe the Hadamard product (via Weierstrass factorization) does globally converge - see my comment above.
You are going to need a bit of knowledge about complex analysis before you can really follow the answer, but if you start with a function defined as a series, it is frequently possible to extend that function to a much larger part of the complex plane.
For example, if you define $f(x)=1+x+x^2+x^3+...$ then $f$ can be extended to $\mathbb C\setminus \{1\}$ as $g(x)=\frac{1}{1-x}$. Clearly, it is "absurd" to say that $f(2)=-1$, but $g(2)=-1$ makes sense.
The Riemann zeta function is initially defined as a series, but it can be "analytically extended" to $\mathbb C\setminus \{1\}$. The details of this really require complex analysis.
Calculating the non-trivial zeroes of the Riemann zeta function is a whole entire field of mathematics.
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