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I am in a hurry now but let me tell what I think. I believe that in the critical strip and off the real axis $\prod_p (1-p^{-s})$ does not converge to any complex number (including zero). Using a similar idea as in my response to your related earlier question, this boils down to the fact that for any nonzero constants $c_1,\dots,c_n$ the sum $\sum_{m=1}^n c_m \sum_{p\in P}p^{-ms}$ oscillates wildly as $P\to\infty$. I dont's see this immediately, but I believe what happens is that the oscillation (or divergence) behavior of the inner sum depends heavily on $m$. More precisely, I believe that for $m=1$ you get a much wilder behavior than for the rest $m>1$. So altogether the above double sum inherits the behavior of $m=1$, namely the existence of very large partial sums for infinitely many $P$'s, and this prevents convergence or a tendency to pointing in special directions. I apologize if this is too vague, but certainly more than a comment.

EDIT 1: David Speyer showed that in the critical strip $\prod_p (1-p^{-s})$ does not converge to any nonzero complex number, see here. I believe that my approach above can also be made to work and yield more information. Perhaps the Riemann Hypothesis can be of great assistance here as $\sum_{p\in P}p^{-s}$ is very subtle. Note that for $\mathrm{Re}(s)=1$ and $s\neq 1$ the Euler product does converge to $\zeta(s)$, see Section 3.15 in Titchmarsh: The Theory of the Riemann Zeta-function.

EDIT 2: In my response to this question, I outline the proof that, assuming the Riemann Hypothesis, the partial products of $\prod_p (1-p^{-s})$ get arbitrary close to $0$ and $\infty$, at least for $\frac{1}{2}<\mathrm{Re}(s)<1$. I don't see any fundamental difficulty in extending this to $\mathrm{Re}(s)=\frac{1}{2}$.

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I am in a hurry now but let me tell what I think. I believe that in the critical strip and off the real axis $\prod_p (1-p^{-s})$ does not converge to any complex number (including zero). Using a similar idea as in my response to your related earlier question, this boils down to the fact that for any nonzero constants $c_1,\dots,c_n$ the sum $\sum_{m=1}^n c_m \sum_{p\in P}p^{-ms}$ oscillates wildly as $P\to\infty$. I dont's see this immediately, but I believe what happens is that the oscillation (or divergence) behavior of the inner sum depends heavily on $m$. More precisely, I believe that for $m=1$ you get a much wilder behavior than for the rest $m>1$. So altogether the above double sum inherits the behavior of $m=1$, namely the existence of very large partial sums for infinitely many $P$'s, and this prevents convergence or a tendency to pointing in special directions. I apologize if this is too vague, but certainly more than a comment.

EDIT: David Speyer showed that in the critical strip $\prod_p (1-p^{-s})$ does not converge to any nonzero complex number, see here. I believe that my approach above can also be made to work and yield more information. Perhaps the Riemann Hypothesis can be of great assistance here as $\sum_{p\in P}p^{-s}$ is very subtle. Note that for $\mathrm{Re}(s)=1$ and $s\neq 1$ the Euler product does converge to $\zeta(s)$, see Section 3.15 in Titchmarsh: The Theory of the Riemann Zeta-function.

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I am in a hurry now but let me tell what I think. I believe that in the critical strip and off the real axis $\prod_p (1-p^{-s})$ does not converge to any complex number (including zero). Using a similar idea as in my response to your related earlier question, this boils down to the fact that for any nonzero constants $c_1,\dots,c_n$ the sum $\sum_{m=1}^n c_m \sum_{p\in P}p^{-ms}$ oscillates wildly as $P\to\infty$. I dont's see this immediately, but I believe what happens is that the oscillation (or divergence) behavior of the inner sum depends heavily on $m$. More precisely, I believe that for $m=1$ you get a much wilder behavior than for the rest $m>1$. So altogether the above double sum inherits the behavior of $m=1$, namely the existence of very large partial sums for infinitely many $P$'s, and this prevents convergence or a tendency to pointing in special directions. I apologize if this is too vague, but certainly more than a comment.