To evaluate that limit, we can expand each function in a Laurent series at $s=0$ and use the following 3 facts about the Hurwitz zeta function:
$$ \zeta(-s,a) = \zeta(-s,a+1) + a^{s} \tag{1}$$
$$ \zeta'(-s,a) = \zeta'(-s,a+1) -a^{s} \log(a) $$
$$\zeta(-n, a) = -\frac{B_{n+1}(a)}{n+1} \ , \ n \in\mathbb{N} \tag{2}$$
Doing so, we get
$$z - z \log z - \frac{\gamma z^{2}}{2} + \lim_{s \to 0^{+}} \Big[ - \Gamma(s-1) \zeta(s-1,z+1) -z \Gamma(s) \zeta(s) + \frac{z^{2}}{2} \Gamma(s+1) \zeta(s+1)$$
$$+ \Gamma(s-1) \zeta(s-1) \Big]$$
$$ = z - z \log z - \frac{\gamma z^{2}}{2}$$
$$ + \lim_{s \to 0^{+}} \Bigg[-\Big(-\frac{1}{s} + \gamma -1 + \mathcal{O}(s) \Big) \Big( -\frac{z^{2}}{2}+\frac{z}{2}-\frac{1}{12}-z + \zeta'(-1,z)s + z \log z \ s + \mathcal{O}(s^{2}) \Big)$$
$$ - z \Big( \frac{1}{s} - \gamma + \mathcal{O}(s) \Big) \Big( - \frac{1}{2} - \frac{\log (2 \pi)}{2} s + \mathcal{O}(s^{2}) \Big) + \frac{z^{2}}{2} \Big(1- \gamma s + \mathcal{O}(s^{2}) \Big) \Big( \frac{1}{s} + \gamma + \mathcal{O} (s) \Big) $$
$$ + \Big(- \frac{1}{s} + \gamma -1 + \mathcal{O} (s) \Big) \Big( - \frac{1}{12} + \zeta'(-1) s + \mathcal{O}(s^{2}) \Big) \Bigg] $$
$$ = z - z \log z - \frac{\gamma z^{2}}{2}$$
$$ + \lim_{s \to 0^{+}} \Big[\zeta'(-1,z) + z \log z + \frac{\gamma z^{2}}{2} - \frac{z^{2}}{2} + \frac{z}{2} - z + \frac{z \log(2 \pi)}{2} - \zeta(-1)+ \mathcal{O}(s) \Big] $$
$$ = \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} -\zeta'(-1)+ \zeta'(-1,z) $$
$ $
$(1)$ http://dlmf.nist.gov/25.11 (25.11.3)
$(2)$ http://mathworld.wolfram.com/HurwitzZetaFunction.html (9)
Another way to show $$\int_{0}^{1} \log \Gamma(x+ \alpha) \, dx = \int_{0}^{1} \log \Gamma(x) \, dx + \alpha \log \alpha - \alpha $$
is to rewrite the integral as
$$ \begin{align} \int_{0}^{1} \log \Gamma (x+\alpha) \, dx &= \int_{\alpha}^{\alpha+1} \log \Gamma(u) \, du \\ &= \int_{0}^{1} \log \Gamma (u) \, du + \int_{1}^{\alpha+1} \log \Gamma (u) \, du - \int_{0}^{\alpha} \log \Gamma (u) \, du \\ &= \int_{0}^{1} \log \Gamma (u) \, du + \int_{0}^{\alpha} \log \Gamma (w+1) \, dw - \int_{0}^{\alpha} \log \Gamma (u) \, du \end{align}$$
and then combine the 2nd and 3rd integrals and use the functional equation $\frac{\Gamma(x+1)}{\Gamma (x)} = x.$
Best Answer
Note that
$$ \begin{align*}I &= \int_0^1 \log \Gamma(x+1) \; dx = \int_0^1 \log \left(x \Gamma(x) \right)\; dx = \int_0^1 \log(x)dx+\int_0^1 \log\Gamma(x) \; dx \\ &=-1+\int_0^1 \log\Gamma(x) \; dx \tag{1} \end{align*}$$
On the substitution $x\mapsto 1-x$, this becomes
$$ I=-1+\int_0^1 \log\Gamma(1-x) \; dx \tag{2} $$
Averaging $(1)$ and $(2)$,
$$ \begin{align*} I &= -1+\frac{1}{2}\int_0^1 \log\left( \Gamma(x)\Gamma(1-x) \right)\; dx \\ &= -1+\frac{1}{2}\int_0^1 \log \frac{\pi}{\sin(\pi x)}dx \tag{3}\\ &= -1+\frac{1}{2}\log(\pi)-\frac{1}{2}\int_0^1 \log(\sin \pi x) dx \\ &= -1+\frac{1}{2}\log(\pi)-\frac{1}{2\pi}\int_0^\pi \log(\sin x)dx \\ &= -1+\frac{1}{2}\log(\pi)+\frac{1}{2}\log(2) \tag{4}\\ &= \log\sqrt{2\pi}-1 \end{align*} $$
Equation $(3)$ follows from the reflection formula of the Gamma Function. To get equation $(4)$, I used the well known result:
$$\int_0^\pi \log(\sin x)dx=-\pi \log(2)$$