The $\chi(s)$ function in the Franca-LeClair paper at page 34 is stated as:
$$\chi(s)=\pi^{-s/2}\Gamma(s/2)\zeta(s) \;\;\;\;\;\;\;\;(1)$$
and it satisfies the functional equation:
$$\chi(s)=\chi(1-s)\;\;\;\;\;\;\;\;(2)$$
What expression $\alpha(s,c)$ in $(3)$ below, completes the Dirichlet generating function $\zeta(s+c-1)$ where $c$ is a constant?
$$\chi_c(s)= \alpha(s,c) \zeta(s+c-1)\;\;\;\;\;\;\;\;(3)$$
so that it also satisfies a functional equation:
$$\chi_c(s)=\chi_c(1-s)\;\;\;\;\;\;\;\;(4)$$
Ultimately I would like to see the completed function for:
$$\frac{\zeta(s)\zeta(c)}{\zeta(s+c-1)} \;\;\;\;\;\;\;\;(5)$$
and run through the Franca LeClair derivation of the asymptotic and see if it pushes the asymptotic closer to the actual zeta zeros in the limit $c \rightarrow 1$.
You shouldn't expect nice symmetry in the functional equation when you shift (or rather, you should expect to the symmetry to shift too). In particular, if you shift, then your function $\alpha(s,c)$ will include the Riemann zeta function in it somehow, so you're not really learning anything useful.
The standard notation for \[\pi^{-s/2} \Gamma\left(\frac{s}{2}\right) \zeta(s)\] is $\Lambda(s)$ or $\xi(s)$, not $\chi(s)$. From the functional equation, we have that \[\Lambda(s) = \Lambda(1 - s).\] If we replace $s$ with $s + c - 1$, then we see that \[\Lambda(s + c - 1) = \Lambda(2 - s - c).\] This is the correct functional equation; it relates $\zeta(s + c - 1)$ to $\zeta(2 - s - c)$. If you wish to relate $\zeta(s + c - 1)$ to $\zeta(c - s)$, the relation is going to still depend on the zeta function.
Suppose we wish to find a function $\alpha(s,c)$ such that \[\alpha(s,c) \zeta(s + c - 1) = \alpha(1 - s,c) \zeta(c - s).\] Well, we have that \[\Lambda(s + c - 1) = \Lambda(2 - s - c)\] and that \[\Lambda(c - s) = \Lambda(1 + s - c).\] That is, \[\pi^{-(s + c - 1)/2} \Gamma\left(\frac{s + c - 1}{2}\right) \zeta(s + c - 1) = \pi^{-(2 - s - c)/2} \Gamma\left(\frac{2 - s - c}{2}\right) \zeta(2 - s - c)\] and \[\pi^{-(c - s)/2} \Gamma\left(\frac{c - s}{2}\right) \zeta(c - s) = \pi^{-(1 + s - c)/2} \Gamma\left(\frac{1 + s - c}{2}\right) \zeta(1 + s - c).\] Combining these two equations, we find that \[\frac{\zeta(s + c - 1)}{\zeta(c - s)} = \frac{\pi^{-(1 - s)}}{\pi^{-s}} \frac{\Gamma\left(\frac{1 - c + (1 - s)}{2}\right) \Gamma\left(\frac{c - (1 - s)}{2}\right)^{-1}}{\Gamma\left(\frac{1 - c + s}{2}\right) \Gamma\left(\frac{c - s}{2}\right)^{-1}} \frac{\zeta(1 - c + (1 - s))}{\zeta(1 - c + s)}. \] So \[\alpha(s,c) = \pi^{-s} \frac{\Gamma\left(\frac{1 - c + s}{2}\right)}{\Gamma\left(\frac{c - s}{2}\right)} \zeta(1 - c + s).\]