It appears that for all $\text{Re}(z)>0$, we have
$$\pi - i \text{Ci}(2\pi-iz)+i \text{Ci}(-2\pi+iz) = 0$$
where
$$\text{Ci}(z) = -\int_z^\infty \frac{\cos(t)}{t} dt$$
Mathematica doesn't simplify this to zero. Graphing it appears to produce the zero function, everywhere I look in $\mathbb{C}$. I have never seen this property before. Does anyone know if it is true and/or where a reference would be? Considering it's simplicity, I would have thought this would be on wikipedia, but didn't see it on there.
Unfortunately, your formula is only correct for $z \in \mathbb{R}_{0}^{+}$. But we can quickly extend it for negative $z$ as well, since: $$ \begin{align*} \pi - \operatorname{Ci}\left( 2 \cdot \pi - z \cdot i \right) \cdot i + \operatorname{Ci}\left( -2 \cdot \pi + z \cdot i \right) \cdot i &= \begin{cases} 0,\, &\text{if}\, z \in \mathbb{R}_{0}^{+}\\ 2 \cdot \pi,\, &\text{if}\, z \in \mathbb{R}^{-}\\ \end{cases}\\ \pi - \operatorname{Ci}\left( 2 \cdot \pi - z \cdot i \right) \cdot i + \operatorname{Ci}\left( -2 \cdot \pi + z \cdot i \right) \cdot i &= 2 \cdot \pi \cdot \theta\left( -x \right)\\ \end{align*} $$ where $\theta\left( \cdot \right)$ is the unit step function aka the Heaviside step function.
Why is that? For the proof we can first state that the equation should result in a constant $c$. And we know $\frac{\operatorname{d}\text{costant}}{\operatorname{d}z} = 0 \wedge \frac{\operatorname{d}\operatorname{Ci}\left( z \right)}{\operatorname{d}z} = \frac{\cos\left( z \right)}{z}$, so the derivative of the reflection formula should also be $0$ if it is constant per interval - wich is ture as you can see here.
I got the claim that $\pi - \operatorname{Ci}\left( 2 \cdot \pi - z \cdot i \right) \cdot i + \operatorname{Ci}\left( -2 \cdot \pi + z \cdot i \right) \cdot i = 2 \cdot \pi \cdot \theta\left( -x \right)$ holds from the graphs. Let's assume that's it, we should be able to just solve for $z$ and get that. But why make it so difficult? We already know that $+\pi$ doesn't change anything about the interval distributions and $\operatorname{Ci}\left( \pm x \cdot i \mp 2 \cdot \pi \right) + \operatorname{Ci}\left( \mp x \cdot i \pm 2 \cdot \pi \right)$ is not continuous only at $x = 0$ (when $x \in \mathbb{R}$) aka it should only have at least two intervals give in which the function is non-equal. So there is a constant for negative $z$ and one for positive $z$. As you have already observed, a value for positive $z$ is $0$ and since it is constant it must be so for every single one else when $z > 0$. Doing the same for negative $z$ ($z < 0$) as I did in my comment we get $2 \cdot \pi$ which is the solution from before aka we derived the formula.