Can we combine two reactions and then calculate the equilibrium concentrations?
Karsten's insight is correct. Your mistake is in assuming that you have either $\ce{N2O5}$ or $\ce{N2O}$ and $\ce{O2}$.

There are four concentrations of interest: $\ce{[N2O5]}$, $\ce{[N2O3]}$, $\ce{[N2O]}$, and $\ce{[O2]}$.

How can you relate these variables?

Well, you can apply conservation of matter in terms of oxygen and nitrogen.

For nitrogen:

$$2\ce{[N2O5]} + 2\ce{[N2O3]} + 2\ce{[N2O]} = 2\times(4\ \mathrm{M})\tag{1}$$

For oxygen:

$$5\ce{[N2O5]} + 3\ce{[N2O3]} + \ce{[N2O]} + 2 \ce{[O2]} = 5\times(4\ \mathrm{M})\tag{2}$$

Then, there are the equilibrium expressions:

$$K_{1}=4.5=\frac{\ce{[N2O3][O2]}}{\ce{[N2O5]}}\tag{3}$$

$$K_{2}=\frac{\ce{[N2O][O2]}}{\ce{[N2O3]}}\tag{4}$$

If you substitute $\ce{[O2]} = 4.5\ \mathrm{M}$ into (2), (3), and (4), you end up with:

$$2\ce{[N2O5]} + 2\ce{[N2O3]} + 2\ce{[N2O]} = 2\times(4\ \mathrm{M})\tag{1}$$
$$5\ce{[N2O5]} + 3\ce{[N2O3]} + \ce{[N2O]} + 2\times(4.5\ \mathrm{M}) = 5\times(4\ \mathrm{M})\tag{2'}$$
$$4.5=\frac{\ce{[N2O3]\times(4.5\ \mathrm{M})}}{\ce{[N2O5]}}\tag{3'}$$
$$K_{2}=\frac{\ce{[N2O]\times(4.5\ \mathrm{M})}}{\ce{[N2O3]}}\tag{4'}$$

That's 4 equations and 4 unknowns, and it's all algebra from there.

Assuming I did that correctly, I got $$K_{2} = \frac{45}{7}$$