Can we combine two reactions and then calculate the equilibrium concentrations? At a certain temperature, $\ce{N2O5}$ dissociates as:
  \begin{align}
\ce{N2O5 (g) &amp;&lt;=&gt; N2O3 (g) + O2 (g)} &amp; K_1 &amp;= 4.5
\end{align} 
  At the same time, $\ce{N2O3}$ also dissociates as:
  \begin{align}
\ce{N2O3 (g) &amp;&lt;=&gt; N2O (g) + O2 (g)}  &amp; K_2 &amp;= \mathrm{?}
\end{align} 
  
  When 4 moles of $\ce{N2O5}$ are heated in a flask of $\pu{1 L}$ volume, it has been observed that the concentration of $\ce{O2}$ is $\pu{4.5 M}$ at
  equilibrium. What are the equilibrium concentrations of the other products?


The combined reaction:

$$\ce{ N2O5(g) &lt;=&gt; N2O(g)  + 2O2(g)}$$
$$K_\text{combined} = K_1 \times K_2 = 4.5 K_2$$

There are 4 moles of $\ce{N2O5}$ initially. If $x$ moles of $\ce{N2O5}$ dissociate to give $x$ moles of $\ce{N2O}$ and $2x$  moles of $\ce{O2}$, then plugging in the values in the $K_c$ expression for this reaction we get 
$$4.5 K_2 = \frac{4x^3}{4-x},$$ 
where 
$$x = \frac{4.5}{2 \pu{mol}}.$$

Solving for $K_2$, I get $K_2=\frac{81}{14}$, which is incorrect.

Is it allowed to first combine the reactions and then use the combined reaction?
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 $$K_{2} = \frac{45}{7}$$