There is no good reason to take the imaginary part of the zeros or to keep only those in $\Re(s)\ge 1/2$.
From the residue theorem we get for $\Re(x) >0$ the holomorphic function
$$F(x)=\sum_{\Im(\rho)>0} e^{i\rho x}=\frac1{2i\pi} \int_{-1+i\infty}^{-1}+\int_{-1}^2+\int_2^{2+i\infty} (\frac{\zeta'(s)}{\zeta(s)}+\frac1{s-1})e^{isx}ds$$
$$=A(x)+B(x)+C(x)$$
Where the first term $$A(x)= \frac1{2i\pi}\int_{-1}^2(\frac{\zeta'(s)}{\zeta(s)}+\frac1{s-1})e^{isx}ds$$
is entire,
The second term depends only on the functional equation
$$B(x)=\frac1{2i\pi}\int_2^{2+i\infty} \frac{e^{isx}}{s-1}ds$$
$$+\frac1{2i\pi} \int_{-1+i\infty}^{-1}(\log(2\pi)+\frac{\pi \cos(\pi s/2)}{2\sin(\pi s/2)}-\frac{\Gamma'(1-s)}{\Gamma(1-s)}+\frac1{s-1})e^{isx}ds$$
And
$$C(x)=\frac1{2i\pi}\int_0^{i\infty} (\frac{\zeta'(2+s)}{\zeta(2+s)}e^{i(2+s)x}+\frac{\zeta'(2-s)}{\zeta(2-s)} e^{i(s-1)x})ds$$
$$ = \frac{1}{2i\pi} \sum_{n\ge 1}\frac{\Lambda(n)}{n^2}(\frac{e^{2ix} }{ix-\log n}+\frac{e^{-ix}}{ix+\log n})$$
$-2\pi e^{-ix/2}C(x)$ is meromorphic on the whole complex plane with simple poles at the $\pm i\log n$ of residue $\frac{\Lambda(n)}{n^{1/2}}$
In other words $F(x)$ is much simpler than it seems.
