Does the Euler product for the Dirichlet $\beta$-function converge for all $\Re(s)>\frac12$?

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The Dirichlet $\beta$-function is defined for $\Re(s)>0$ as:

$$\beta(s) = \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^s}$$

It has the following Euler product (I used that Dirichlet character $\chi_{4}(p)=\sin\left(\frac{p \,\pi}{2}\right)$):

$$\prod_p \bigg(\frac {p^s}{p^s-\sin\left(\frac{p \,\pi}{2}\right)} \bigg)$$

Numerical evidence suggests that this Euler product also (slowly) converges for values $\Re(s)>\frac12$.

Does convergence in the domain $\frac12 < \Re(s) \le 1$ indeed occur? If so, could this be proven (I guess a proof would also imply that all complex zeros of $\beta(s)$ must reside on the critical line)?

P.S.:

Just to share that by multiplying the Leibniz formula for $\pi$ ($\beta(1)=\frac{\pi}{4}$) and this formula (35), gives the very elegant relationship:

$$\prod_p \bigg(\frac{p-\sin\left(\frac{p \,\pi}{2}\right)}{{p+\sin\left(\frac{p \,\pi}{2}\right)}} \bigg)=2$$

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Since the product (over primes) for Dirichlet L-series is valid only for $\Re{(s)}>1$, as mentioned in all textbooks on this subject, the answer to Agno initial question is that not even the `easiest' case $s=1$ is proved for $\prod_p \bigg(\frac {p^s}{p^s-\sin\left(\frac{p \,\pi}{2}\right)} \bigg)$. I have investigated these products for $1/2 < \Re{(s)} \le 1$ for a while and also detected a slow numerical convergence, but this was just a conclusion on experimental computations, nothing such as a formal proof.

P.S.: Could someone (maybe Agno, himself, or Reuns) provide a reference in literature (or internet) for the formal proof of "the equivalence between the zeros and the abscissa of convergence" mentioned by Reuns?