My first attempt:
$$f(x) = \prod_{i=1}^\infty \left(1- \frac x {p_i} \right)$$
If we take a look at the Riemann zeta function:
$$ \zeta(s) = \sum_{n = 1}^\infty \frac 1 {n^s} = \prod_{i = 0}^\infty \left(\frac {1}{1 - p_i^{-s}} \right) = \prod_{i = 0}^\infty \left(1 - \frac {1}{p_i^{s}} \right)^{-1} $$
$$f(1) = \frac 1 {\zeta(1)} = 0$$
By $f$'s construction, it should only contain $0$ factors at prime $x$s, which $1$ is not. Therefore, the only reason $f$ should be $0$ at $1$ is that the product converges to $0$ as $i \to \infty$.
My second attempt:
$$g(x) = \prod_{i=1}^\infty \left(1- \frac {x^2} {p_i^2} \right)$$
However, I have no real idea how to show that this second attempt does not converge to $0$ on non-prime values of $x$. How do I show that either $g(x)$ converges to $0$ at some non-prime point $c$, or show that $g(x)$ is only $0$ for prime values $x$?
A possible construction is $$ \prod \left(\bigg(1-\frac{x}{p_i}\bigg) e^{\frac{x}{p_i}}\right). $$
Say, for $p_i>2|x|$, $$ \log \left(\bigg(1-\frac{x}{p_i}\bigg) e^{\frac{x}{p_i}}\right) = O\left(\frac{|x|^2}{p_i^2}\right), $$ This ensures that the tail of the product converges and the limit is nonzero.
See also Weierstrass' factorization theorem.