So a very important formula proven by Euler is that
is equal to
Of course these formulas give you the same value when they reach infinity, but my question is that say $s=1$. What would be the difference in values between these two formulas if n and p didn't go to infinity and instead stopped at any finite would they always be very close or would there difference become greater and greater.
The short answer is that the two values $$ \sum_{n \leq P} \frac{1}{n^s}, \qquad \prod_{p \leq P} \left( 1 - \frac{1}{p^s}\right)^{-1}$$ are very different for all $s > 1$.
But there is something that can be said. For the rest of this answer, I will instead answer the following question:
I think this is at the heart of your question. In answering this question, I will inadvertently provide an answer to your original question, which (seems to) consider $\mathcal{N} = [1, P]\cap\mathbb{Z}$.
A single factor in the Euler product looks like $$ \frac{1}{1 - \frac{1}{p^s}}.$$ This can be expanded as a geometric series into $$ \frac{1}{1 - \frac{1}{p^s}} = 1 + p^{-s} + p^{-2s} + p^{-3s} \cdots$$ This converges absolutely for $s > 0$. If we take this product with the factor corresponding to another prime $q$, then we get $$ \frac{1}{1 - \frac{1}{p^s}} \cdot \frac{1}{1 - \frac{1}{q^s}} = (1 + p^{-s} + p^{-2s} + \cdots)(1 + q^{-s} + q^{-2s} + \cdots).$$ In the expansion of the terms on the right, we can form $n^{-s}$ for exactly those integers $n$ of the form $n = p^j q^k$, namely by multiplying the $j$th term in the first sum by the $k$th term in the second. So $$\frac{1}{1 - \frac{1}{p^s}} \cdot \frac{1}{1 - \frac{1}{q^s}} = \sum_{n \in \mathcal{N}_{p,q}} \frac{1}{n^s},$$ where $\mathcal{N}_{p,q}$ denotes exactly those integers formed from the primes $p$ and $q$.
More generally but in the same vein, we have $$ \prod_{p \in \mathcal{P}} \left( 1 - \frac{1}{p^s}\right)^{-1} = \sum_{n \in \mathcal{N}_{\mathcal{P}}} \frac{1}{n^s},$$ where $\mathcal{P}$ is a collection of primes and $\mathcal{N}_{\mathcal{P}}$ denotes the set of integers formed from products of primes in $\mathcal{P}$.
To answer my question, the right $\mathcal{N}$ is the set of integers formed from the primes up to $P$, in which case there is an exact equality. Note that this is an infinite sum!
This also means that $$ \prod_{p \leq P} \left(1 - \frac{1}{p^s}\right)^{-1} - \sum_{n \leq P} \frac{1}{n^s} = \sum_{\substack{n > P \\ n \in \mathcal{N}_{p \leq P}}} \frac{1}{n^s} - \sum_{\substack{n \leq P \\ n \not \in \mathcal{N}_{p \leq P}}} \frac{1}{n^s}.$$