The answer to your monotonicity conjecture is affirmative. It admits a somewhat sneaky proof and allows us to conclude something about the Poisson distribution in the
process. This is what we explore below.
The picture
The question asks whether the cdfs decrease pointwise as $n$ increases
for each $x$ to the left of the vertical line $x=1$.
The math
Let's start with a restatement of the question. First, if $\renewcommand{\Pr}{\mathbb P}\newcommand{\d}{\mathrm d}T \sim
\mathrm{Exp}(1)$, then it's easy to see that $T/n \sim
\mathrm{Exp}(n)$. So, we can rewrite a $\Gamma(n,n)$ random variable
$S_n$ as $S_n = (T_1 + \cdots + T_n)/n$ where $T_i$ is a sequence of
iid $\mathrm{Exp}(1)$ random variables. Instead of asking whether the
cdf decreases pointwise as a function of $n$ for each $0 < x < 1$, we
can ask whether the survival distribution increases
pointwise. Hence, we have the following equivalent question.
Equivalent question: Let $T_1, T_2,\ldots$ be an iid sequence of $\mathrm{Exp}(1)$ random variables and let $S_n = T_1 + \dots + T_n$. Then, is it true that, for all fixed $x \in (0,1)$, $$ \Pr( S_{n+1} > (n+1) x ) \geq \Pr(S_n > n x) \,? $$
Proof. The key idea here is to divide and conquer. We need to somehow compare the events $\{S_{n+1} > (n+1)x\}$ and $\{S_n > nx\}$. To that end, notice that
$$
\Pr(S_{n+1} > (n+1)x) = \Pr(S_{n+1} > n x) - \Pr(nx \leq S_{n+1} \leq (n+1)x) \>.
$$
We are halfway there. To deal with the remaining part, observe that
$$
\{S_{n+1} > n x\} = \{S_n > n x\} \cup \{S_n \leq n x, T_{n+1} > nx - S_n \} \>,
$$
and the two events on the right-hand size are disjoint. So,
$$
\Pr(S_{n+1} > (n+1)x) = \Pr(S_n > n x) + \Pr(S_n \leq nx, T_{n+1} > nx - S_n) - \Pr(nx \leq S_{n+1} \leq (n+1)x) \>.
$$
If we can show the second probability is greater than the third, we are done. To do this, we'll first find an explicit value for the second probability and then find an upper bound for the third.
(Second term.) $S_n$ and $T_{n+1}$ are independent, so to find $\Pr(S_n \leq nx, T_{n+1} > nx - S_n)$, we need only integrate over the
region of interest. This is
$$
\Pr(S_n \leq nx, T_{n+1} > nx - S_n) = \int_0^{nx} \int_{nx-s}^\infty
e^{-t} \frac{s^{n-1}e^{-s}}{(n-1)!} \, \d t \, \d s =
\frac{e^{-nx}(nx)^n}{n!} \>.
$$
(Third term.) This step is more involved, but still employs only elementary tools. Using the density of $S_{n+1}$, we have
$$\newcommand{\d}{\mathrm d}
\Pr(nx \leq S_{n+1} \leq (n+1)x )= \int_{nx}^{(n+1)x} \frac{u^n
e^{-u}}{n!} \,d u = \frac{e^{-nx} (nx)^n}{n!} x \int_0^1 (1+v/n)^n
e^{-xv} \d v
$$
where we've obtained the integral on the far right by using the substitution $v =
(u-nx)/x$. Now $(1+v/n)^n < e^v$ for all $n$ and so
$$
\int_0^1 (1+v/n)^n e^{-xv} \d v < \int_0^1 e^{(1-x) v} \d v =
\frac{e^{1-x} - 1}{1-x} \leq \frac{1}{x} \>,
$$
where the last step follows from the fact that $e^{1-x} < x^{-1}$ for
all $x \in (0,1)$. So,
$$
\Pr(nx \leq S_{n+1} \leq (n+1)x ) < \frac{e^{-nx}(nx)^n}{n!} \>.
$$
But, this last quantity is the same as the second term! Putting the three pieces together, we get
$$
\Pr(S_{n+1} > (n+1)x ) > \Pr(S_n > n x) \>,
$$
for each $x \in (0,1)$, which is what we set out to prove.
A consequence
The appearance of the term $\frac{e^{-nx}(nx)^n}{n!}$ strongly hints
that there is a relation to the Poisson hiding in here. Indeed, the
proof implies the following.
Proposition: Let $X_{n\lambda} \sim \mathrm{Poisson}(n\lambda)$ for $\lambda \in (0,1)$. Then $$ \Pr(X_{n\lambda} < n) \geq e^{-\lambda} \>. $$
Indeed, this was the original context in which I recently gave an equivalent
proof in this MathOverflow answer.
