Convergence of Riemann's Product representation of Xi

You need to group the complex conjugates pairs of non-trivial zeros together

$$2sum_{rho} log(1-frac{s}{rho})= 2sum_{Im(rho)le 2|s|} log(1-frac{s}{rho})+ 2sum_{Im(rho)> 2|s|}log(1-frac{s}{rho})+log(1-frac{s}{overline{rho}})$$ $$=2sum_{Im(rho)le 2|s|} log(1-frac{s}{rho})+ 2sum_{Im(rho)> 2|s|}O(frac{|s|+|s^2|}{Im(rho)^2})tag{1}$$ Due to the density of zeros theorem $$# {rho, |Im(rho)|<T}=O(Tlog T)$$ $(1)$ converges locally uniformly, with a rate of convergence similar to $sum_{n>2|s|log |s|} frac{(|s|+|s^2|) log^2 n}{n^2 }$.

As a comment says, this is "standard" textbook material nowadays, but I'm guessing you want a more historical perspective from Riemann's point of view.

The general theory of the Hadamard product (for entire functions of order 1) obviously wasn't available, but one really only needs linear exponential factors, so Riemann essentially did this ad hoc.

The relevant part is in the middle of page 139 of [1].

The translation is given as

If one denotes by $alpha$ all the roots of the equation $xi(alpha)$ = 0, one can express $log xi(t)$ as $$sum_alpha log(1-t^2/alpha^2)+logxi(0)$$ for, since the density of the roots of the quantity $t$ grows with $t$ only as $log t/2pi$, it follows that this expression converges [the important point]... thus it differs from $logxi(t)$ by a function of $t^2$... This difference is consequently a constant, whose value can be determined through setting $t = 0$.

So in other words, I think a fair answer to your question is that Riemann instead considers $$xi(s)=xi(0)prod_rho (1-s^2/rho^2),$$ which softens the analytic difficulties compared to $xi(s)=prod_rho (1-s/rho)$.

[1] https://upload.wikimedia.org/wikipedia/commons/c/cb/Ueber_die_Anzahl_der_Primzahlen_unter_einer_gegebenen_Gr%C3%B6sse.pdf