There is no key point, only tedious calculation. It is merely a coincidence. A theory can either be renormalisable, or non-renormalisable; and both scenarios are in principle conceivable. Power-counting renormalisability usually offers a strong hint towards deciding whether the theory is renormalisable or not, but this is not an infallible test1.
Quantum gravity is not power-counting renormalisable, so in principle you should suspect that the theory is not renormalisable. But it may very well be the case that some miraculous cancellation of divergences comes to the rescue and makes the theory renormalisable after all. Indeed, the theory might contain some hidden symmetry that controls the possible divergences. The fact that the first few loop orders turn out to be finite is no proof that they are finite to any order. You either prove that they are (which is a very non-trivial task), or prove that they aren't (by finding an explicit counter-example).
In the case of QG, the first order happens to be finite. There are other examples of theories that are one-loop finite but not to higher orders (e.g., naïve massive Yang-Mills, cf. this PSE post). One does not really need to explain the one-loop finiteness: it just sometimes happens, with no deep reason behind it. It turns out that in QG one may partially explain this phenomenon on grounds of the metric-independence of the Euler-Poincaré characteristic. Quoting DeWitt,
Because the metric independence of the Euler-Poincaré characteristic[2], terms quadratic in the full Riemann tensor, in the counter-term needed to cancel the pole term [in the one loop effective action], can be replaced by terms quadratic in the Ricci tensor and in the curvature scalar. The counter-term, thus modified, has the form
\begin{equation}
\begin{aligned}
\Delta S&=\frac{1}{16\pi^2}\frac{1}{d-4}\int g^{1/2}\left(-\frac{429}{36}R^2+\frac{187}{90}R_{\mu\nu}R^{\mu\nu}\right)\mathrm d^4x\\
&=\int\frac{\delta S}{\delta g_{\mu\nu}}A_{\mu\nu}\ \mathrm d^4x
\end{aligned}\tag{35.170}
\end{equation}
where
\begin{equation}
A_{\mu\nu}=-\frac{1}{16\pi^2\mu^2}\frac{1}{d-4}\left(\frac{187}{180}R_{\mu\nu}+\frac{979}{180}g_{\mu\nu}R\right)
\end{equation}
Equation $(35.170)$ has exactly the form $(25.90)$. As explained in chapter 25 the presence or absence of the counter-term is therefore irrelevant in the computation of the $S$-matrix, and pure quantum gravity is one-loop finite. This is an accident arising from the existence of the Euler-Poincaré characteristic and does not occur in higher orders.
(Emphasis mine)
For completeness, we sketch the proof of one-loop finiteness of vacuum quantum gravity. We mainly follow 0550-3213(86)90193-8 (§3.1). A simple power-counting analysis reveals that, to one loop, the most general counter-term reads
$$
\Delta S^{(1)}=\int g^{1/2}(c_1R^2+c_2R^{ab}R_{ab}+c_3R^{abcd}R_{abcd})\ \mathrm d^4x
$$
for some (formally divergent) constants $c_{1,2,3}$. The first two terms vanish on-shell (in vacuum), while the third in principle does not. But, using the fact that the Euler-Poincaré characteristic is topological (i.e., its integrand is a total derivative), we may write the $c_3$ term as a function of $R^2$ and $R^{ab}R_{ab}$. This in turns means that
$$
\Delta S^{(1)}\overset{\mathrm{O.S.}}=\int g^{1/2}(\text{total derivative})\ \mathrm d^4x
$$
which proves the one-loop finiteness of quantum gravity (recall that topological terms are invisible to perturbation theory). It is clear that this argument fails in the presence of matter, because the on-shell fields do not satisfy $R^{ab}=0$ anymore, and therefore $\Delta S^{(1)}$ is no longer a total derivative.
In the case of two or more loops, the number of available invariants that can be constructed from the metric, and that may appear as counter-terms, is higher than in the one-loop case. Most of these invariants depend on $R^{abcd}$ rather than $R^{ab}$, and therefore they do not vanish on-shell. In fact, we have
$$
\Delta S^{(2)}\overset{\mathrm{O.S.}}=c_4\int g^{1/2}R^{ab}{}_{cd}R^{cd}{}_{ef}R^{ef}{}_{ab}\ \mathrm d^4x
$$
for some constant $c_4$. Here, there is no identity that relates this combination to a topological term and therefore, unless there is some fortuitous cancellation of divergences that leads to $c_4=0$, the two-loop counter-term Lagrangian is not expected to vanish on-shell. The explicit calculation proves that there is no such cancellation, and therefore quantum gravity is not two-loop finite.
To reiterate, it could have been the case that quantum gravity is renormalisable after all. The one-loop finiteness can be established by simple power-counting arguments, but no such conclusion can be reached for higher loops. Thus, the only thing we can do is to go through the tedious calculation. Once we do this, we find that quantum gravity is not finite. Oh well.
1: Take your favourite renormalisable theory and perform a non-linear field redefinition; the resulting theory has new terms that are not power-counting renormalisable, but the $S$ matrix remains the same (it is finite).
2:
The Euler-Poincaré characteristic in four dimensions reads
$$
\chi_4=\frac{1}{32\pi^2}\int g^{1/2}(R_{abcd}R^{abcd}-4R_{ab}R^{ab}+R^2)\ \mathrm d^4x
$$
