Lets do this as explicitly as possible (I would like to find the constant and error term)
1 The Original Problem
First, consider the identity $\log\left(1-\frac{2}{p}+\frac{1}{p^2}\right) +\log\left(1- \frac{1}{(p-1)^2} \right)=\log\left(1- \frac{2}{p} \right)$. From this, it follows that
$$\sum_{2<p\leq n}\log\left(1-\frac{2}{p}+\frac{1}{p^2}\right) +\sum_{2<p\leq n} \log\left(1- \frac{1}{(p-1)^2} \right) = \sum_{2<p\leq n}\log\left(1-\frac{2}{p}\right)$$
Multiplying by negative one and exponentiating both sides yields
$$\left(\prod_{2<p\leq n}1-\frac{1}{p}\right)^{-2} \cdot \prod_{2<n<p} \left(1- \frac{1}{(p-1)^2} \right)^{-1} = \left(\prod_{2<p\leq n}1-\frac{2}{p}\right)^{-1} $$
Recall $ \Pi_2=\prod_{2<p} \left(1- \frac{1}{(p-1)^2} \right)$ is the Twin Prime Constant, and that the one product on the left hand side converges to the reciprocal of this. It is then by one of Mertens formulas that we know $$\left(\prod_{2<p\leq n}1-\frac{1}{p}\right)^{-1}=\frac{1}{2}e^\gamma \log n + O(1)$$ where $\gamma$ is the Euler Mascheroni Constant. (Specifically this is Theorem 2.7 (e) in Montgomery Multiplicative Number Theory I. Classical Theory) Upon squaring this asymptotic result, we are able to conclude:
$$\left(\prod_{2<p\leq n}1-\frac{2}{p}\right)^{-1} = \frac{1}{4}e^{2\gamma}\Pi_2^{-1} \log^2n + O(\log n)$$
Hope that helps,
Note: The reason I substituted the twin prime constant in for that other product is because it converges very fast in comparison with the error term. I can give more details if desired, but I'll leave it as an exercise.
2 What is best possible?
Can the error term be made better? Yes. It turns out we can make that error term a lot better. By using the Prime Number Theorem we find
$$\prod_{2<p\leq n} \left( 1-\frac{1}{p}\right)^{-1}=\frac{1}{2}e^\gamma \log n + e^{-c\sqrt{\log n}}$$ where $c$ is the constant used in the proof of the Zero Free Region. Since $e^{-\sqrt{\log n}}$ decreases faster than any power of $\log$, we obtain a much better result upon squaring this estimate. Precisely we have:
$$\left(\prod_{2<p\leq n}1-\frac{2}{p}\right)^{-1} = \frac{1}{4}e^{2\gamma}\Pi_2^{-1} \log^2n + O\left( e^{-c\sqrt{\log n}} \right)$$
(again the convergence to $\Pi_2$ is much to rapid to interfere with the error term)
I would be willing to bet that this is the best we can do, and that any better would imply stronger results regarding the error term for $\pi(x)$, the prime counting function.
3 Numerics
Just for fun, the exact constant in front of the $\log^2n$ is: 1.201303. How close is this? Well for:
$n=10$ we get an error of $0.630811$
$n=50$ we get an error of $1.22144$
$n=100$ we get an error of $0.63493$
$n=1000$ we get an error of $0.438602$
$n=10^4$ we get an error of $0.250181$
$n=10^5$ we get an error of $0.096783$
$n=10^6$ we get an error of $0.017807$
Where each time the error is positive. That is the product seems to be slightly larger than the asymptotic (but converging fairly rapidly). However, my intuition tells me it is almost certain (I will not prove it here) that the error term oscillates between negative and positive infinitely often.
4 Under Riemann Hypothesis
If we assume the Riemann Hypothesis, the error term is bounded by $$\frac{C\log^2 x}{\sqrt{x}}$$ for some constant C. By analyzing the above data with numerical methods, the error seems to be best fitted by $\frac{C\log x}{\sqrt {x}}$.
Mertens' third theorem is just the exponentiated version of the second theorem (without the bounds that Mertens proved for his second theorem):
\begin{align}
-\ln\Biggl(\ln n\prod_{p\leqslant n}\biggl(1 - \frac{1}{p}\biggr)\Biggr)
&= -\ln \ln n - \sum_{p\leqslant n} \ln \biggl(1 - \frac{1}{p}\biggr)\\
&= \Biggl(\sum_{p\leqslant n}\frac{1}{p} - \ln \ln n - M\Biggr) + \Biggl(M - \sum_{p\leqslant n} \biggl(\ln\biggl(1-\frac{1}{p}\biggr) + \frac{1}{p}\biggr)\Biggr),
\end{align}
where the first term converges to $0$ by Mertens' second theorem, and the second term converges to $\gamma$ by definition of $M$.
Mertens' bounds in the second theorem and estimates for
$$\sum_{p > n}\biggl(\ln\biggl(1-\frac{1}{p}\biggr)+\frac{1}{p}\biggr)$$
give you bounds for
$$e^\gamma\ln n\prod_{p\leqslant n}\biggl(1-\frac{1}{p}\biggr),\tag{$\ast$}$$
and conversely bounds for that give you bounds for
$$\left\lvert\sum_{p\leqslant n}\frac{1}{p} - \ln \ln n - M\right\rvert,\tag{$\ast\!\ast$}$$
but it is doubtful whether one can directly prove bounds for $(\ast)$ that give you back Mertens' bounds for $(\ast\ast)$.
One can use Mertens' first theorem to derive the second via an integration by parts, Hardy and Wright for example do that, but don't give explicit bounds on $(\ast\ast)$.
For $x > 0$ we define
$$S(x) := \sum_{p\leqslant x} \frac{\ln p}{p}.$$
Mertens' first theorem tells us
$$\lvert S(x) - \ln x\rvert \leqslant 2 + O(x^{-1}),$$
and we can write
$$T(x) := \sum_{p\leqslant x} \frac{1}{p} = \int_{3/2}^x \frac{1}{\ln t}\,dS(t)$$
with a (Riemann/Lebesgue-) Stieltjes integral. Integration by parts yields
\begin{align}
T(x) &= \int_{3/2}^x \frac{1}{\ln t}\,dS(t)\\
&= \frac{S(x)}{\ln x} - \frac{S(3/2)}{\ln \frac{3}{2}} - \int_{3/2}^x S(t)\,d\biggl(\frac{1}{\ln t}\biggr)\\
&= \frac{S(x)}{\ln x} + \int_{3/2}^x \frac{S(t)}{t(\ln t)^2}\,dt\\
&= \frac{S(x)}{\ln x} + \int_{3/2}^x \frac{dt}{t\ln t} + \int_{3/2}^x \frac{S(t) - \ln t}{t(\ln t)^2}\,dt\\
&= \ln \ln x + \underbrace{1 - \ln \ln \frac{3}{2} + \int_{3/2}^\infty \frac{S(t) - \ln t}{t(\ln t)^2}\,dt}_M + \underbrace{\frac{S(x)-\ln x}{\ln x} - \int_x^\infty \frac{S(t)-\ln t}{t(\ln t)^2}\,dt}_{O\bigl(\frac{1}{\ln x}\bigr)}.
\end{align}
I'm not sure, however, whether one can get exactly Mertens' bounds on $(\ast\ast)$ easily from that.
So in a way, Mertens' first theorem is the most powerful, since it implies the others, at least if we don't need explicit bounds for the differences.
Best Answer
See Theorem 6.12 in Pierre Dusart's Estimates of some functions over primes without RH. In particular for all $x \ge 2973$ one has the effective unconditional lower bound: $$\prod_{p\le x}\left(1 - \frac1p\right) > \frac{e^{-\gamma}}{\ln x}\left(1 - \frac{0.2}{\ln^2 x}\right).$$
So, already for $x$ in this range it is true that one can take $c = 1.78666$, which is much smaller than $c=\tfrac95$. A small amount of computation would be all it takes to get the smallest $x$ that works.
Update: it looks like $c=2$ is true for $n\ge 14$, while $c=\tfrac95$ holds for $n \ge 469$.