It is true that $\operatorname{ex}(n; G)/\binom{n}{2}$ is monotone non-increasing. As you've observed, if you have a graph on $n$ vertices of density $\theta$, removing just any vertex doesn't give you a subgraph of density $\theta$. However, it is true that there exists some vertex whose removal gives you a subgraph of density at least $\theta$, as the following "averaging argument" makes precise.
(Throughout the proof, $E(G)$ denotes the edge set of a graph $G$, and $e(G) := \lvert E(G) \rvert$ denotes the number of edges in $G$.)
Proof: Let $G$ be a graph. Let $F$ be a graph on $n$ vertices. Define
$$\theta := \frac{e(F)}{\dbinom{n}{2}}$$
to be the density of $F$.
Now choose $n - 1$ vertices of $F$ uniformly at random, and consider the subgraph $H$ induced by these vertices. There are $n$ choices for $H$, and an edge $uv$ of $F$ is in $H$ as long as both $u$ and $v$ are vertices of $H$. Hence, for all $e \in E(F)$, we have
$$\mathbb{P}\bigl( e \in E(H)\bigr) = \frac{n-2}{n}.$$
This means that the expected number of edges in $H$ is
$$\mathbb{E}\bigl(e(H)\bigr) = \sum_{e \in E(F)} \mathbb{P}\bigl(e \in E(H)\bigr) = \frac{n-2}{n} e(F) = \frac{n-2}{n} \theta\dbinom{n}{2} = \theta\dbinom{n-1}{2}.$$
Because $H$ has $n - 1$ vertices, this means that the expected density of $H$ is $\theta$. Since the average density of $H$ is $\theta$, there must exist a choice of $H$ (call it $H_0$) with density at least $\theta$.
Because $H_0$ is a graph on $n - 1$ vertices, if $\theta > \operatorname{ex}(n - 1; G)/\binom{n - 1}{2}$, then by definition $H_0$ must contain a copy of $G$, which means that and $F$ does as well. We have thus shown that every graph on $n$ vertices of density $\theta > \operatorname{ex}(n - 1; G)/\binom{n - 1}{2}$ must contain a copy of $G$. This implies that
$$\frac{\operatorname{ex}(n; G)}{\dbinom{n}{2}} \leq \frac{\operatorname{ex}(n - 1; G)}{\dbinom{n-1}{2}},$$
as claimed. $\square$
Note: this proof is based on an argument in this survey paper by Peter Keevash (see Section 2).