$\newcommand\Ga\Gamma \newcommand\tp{\tilde p} \newcommand\tq{\tilde q} \newcommand\tr{\tilde r}$First, let us show that $\dfrac{\Ga(s,s-1)}{\Ga(s)}$ decreases in real $s>1$. This is equivalent to $\dfrac{\Ga(s)-\Ga(s,s-1)}{\Ga(s,s-1)}$ being increasing in real $s>1$. Note that \begin{align*} \Ga(s)-\Ga(s,s-1)&=\int_0^{s-1} t^{s-1}e^{-t}\,dt=(s-1)^s\int_0^1 x^{s-1}e^{-(s-1)x}\,dx, \\ \Ga(s,s-1)&=\int_{s-1}^\infty t^{s-1}e^{-t}\,dt=(s-1)^s\int_1^\infty x^{s-1}e^{-(s-1)x}\,dx, \\ \end{align*} So, what we need to show is that \begin{equation} R(u):=\frac{I(u)}{J(u)} \end{equation} is increasing in real $u>0$, where \begin{align*} I(u)&:=\int_0^1 f(x)^u\,dx=\int_0^1 z^u\,p(z)\,dz, \\ J(u)&:=\int_1^\infty f(x)^u\,dx=\int_0^1 z^u\,q(z)\,dz, \end{align*} where $f(x):=xe^{1-x}$, $p(z):=x_1'(z)>0$, $q(z):=-x_2'(z)>0$, $x_1(z)$ is the only root $x\in(0,1)$ of the equation $f(x)=z$ for $z\in(0,1)$, and $x_2(z)$ is the only root $x\in(1,\infty)$ of the equation $f(x)=z$ for $z\in(0,1)$.
To show this, note that \begin{align*} 2J(u)^2R'(u)&=2\int_0^1\int_0^1 dx\,dy\,(xy)^u p(x)q(y)(\ln x-\ln y) \\ &=2\int_0^1\int_0^1 dy\,dx\,(yx)^u p(y)q(x)(\ln y-\ln x) \\ &=\int_0^1\int_0^1 dy\,dx\,(xy)^u [p(x)q(y)-p(y)q(x)] (\ln x-\ln y)\\ &=\int_0^1\int_0^1 dy\,dx\,(xy)^u\,p(y)q(y) [r(x)-r(y)](\ln x-\ln y), \end{align*} where \begin{equation} r:=p/q. \end{equation} So, it remains to show that $r$ is increasing on $(0,1)$.
We have \begin{equation} p=\frac{x_1}{(1-x_1) z},\quad q=\frac{x_2}{(x_2-1) z}, \end{equation} \begin{equation} r'=\frac{x_1(x_2-x_1)(x_1+x_2-2)}{(1-x_1)^3 (x_2-1) x_2 z}. \end{equation}
So, it remains to show that $x_1+x_2-2>0$ or, equivalently, $x_2>2-x_1$ or, equivalently, $f(x_2)<f(2-x_1)$ or, equivalently, $f(x_1)<f(2-x_1)$.
So, it remains to show that $f(x)<f(2-x)$ for $x\in(0,1)$ or, equivalently, $1<h(x):=(2/x-1)e^{2x-2}$ for $x\in(0,1)$, which follows because $h(1)=1$ and $h'(x)=-\dfrac{2 e^{2 x-2} (1-x)^2}{x^2}<0$, so that $h$ is decreasing on $(0,1)$. $\Box$
Let us also prove that $\dfrac{\Ga(s,s)}{\Ga(s)}$ increases in real $s>0$. This proof is similar to the one above. Here in place of $p,q,r=p/q$ we get \begin{equation} \tp:=pe^{-x_1},\quad \tq:=qe^{-x_2},\quad \tr:=\tp/\tq, \end{equation} with \begin{equation} \tr'=\frac{(x_2-x_1) (x_1 x_2-1)}{(1-x_1)^3 (x_2-1) z} \end{equation}
So, it remains to show that $x_1x_2-1<0$ or, equivalently, $x_2<1/x_1$ or, equivalently, $f(x_2)>f(1/x_1)$ or, equivalently, $f(x_1)>f(1/x_1)$.
So, it remains to show that $f(x)>f(1/x)$ for $x\in(0,1)$ or, equivalently, $1<g(x):=x^2e^{1/x-x}$ for $x\in(0,1)$, which follows because $g(1)=1$ and $g'(x)=-(1-x)^2e^{1/x-x}<0$, so that $g$ is decreasing on $(0,1)$. $\Box$
Another way to prove the inequalities $x_1+x_2-2>0$ and $x_1x_2-1<0$, used in the proofs above, is to note that the condition $f(x_1)=f(x_2)$ means that the logarithmic mean of $x_1,x_2$ is $1$, and then use the arithmetic-logarithmic-geometric mean inequality.
The same method can be used to establish monotonicity of the ratio $\dfrac{\Ga(s,s+a)}{\Ga(s)}$ for real $a\notin\{-1,0\}$. For any given real $a$, it will follow that (i) $\dfrac{\Ga(s,s+a)}{\Ga(s)}$ is increasing in real $s>\max(0,-a)$ if the function $$m_a:=(a+1) (x_1 x_2-1)-a (x_1+x_2-2)$$ is $<0$ on $(0,1)$ and (ii) $\dfrac{\Ga(s,s+a)}{\Ga(s)}$ is decreasing in real $s>\max(0,-a)$ if $m_a>0$ on $(0,1)$. So, it follows from the inequalities $x_1 x_2-1<0<x_1+x_2-2$, proved above, that (i) $\dfrac{\Ga(s,s+a)}{\Ga(s)}$ is increasing in real $s>0$ for each real $a\ge0$ and (ii) $\dfrac{\Ga(s,s+a)}{\Ga(s)}$ is decreasing in real $s>-a$ for each real $a\le-1$. It also appears that $\dfrac{\Ga(s,s+a)}{\Ga(s)}$ is decreasing in real $s>-a$ for each real $a\le-1/2$.