Solution:

Observe that the condition is equivalent to saying that every entry in the table has the same parity as the sum of the entries in its row (resp. column).

We claim that upon filling $a_{1}$, $a_{2}$, $b_{1}$, $b_{2}$ there is exactly one way to fill the table. Indeed, we require that

- $a_{3}$ has the same parity as $a_{1} + a_{2}$,
- $b_{3}$ has the same parity as $b_{1} + b_{2}$,
- $c_{1}$ has the same parity as $a_{1} + b_{1}$,
- $c_{2}$ has the same parity as $a_{2} + b_{2}$.

All that's left to do is check that there is exactly one way to fill in the entry $c_{3}$. We require $c_{3}$ to have the same parity as $a_{3} + b_{3}$. But the parity of this is the same as that of $a_{1} + a_{2} + b_{1} + b_{2}$. Similarly, $c_{1} + c_{2}$ has the same parity as $a_{1} + b_{1} + a_{2} + b_{2}$. Hence there is exactly one way to fill in $c_{3}$.

Thus the claim is proved, and the number of ways to fill the entire table is merely the number of ways to fill the $2 \times 2$ upper-left square, which is $2^{4} = 16$ as desired.