Step 1: Build the answer directly instead of subtracting.
Instead of finding the complement (product odd) and subtracting, we can directly count the subsets whose product is even, by using the multiplication principle.
Step 2: Split into 1013 odd numbers and 1013 even numbers.
$S = \{1, \ldots, 2026\}$ contains $1013$ odd numbers and $1013$ even numbers. A subset $M$'s product is even exactly when $M$ contains at least one even number, no matter what odd numbers it also contains.
Step 3: Choose the odd part and the even part independently.
We can build any subset $M$ with product even in two independent choices:
(i) choose any subset of the $1013$ odd numbers, this part is completely free, giving $2^{1013}$ possibilities (including choosing none of the odd numbers),
(ii) choose a non-empty subset of the $1013$ even numbers (it must be non-empty, otherwise there would be no even factor at all), giving $2^{1013} - 1$ possibilities.
Since these two choices are independent and together completely determine $M$, by the multiplication principle the number of subsets with an even product is:
\[ 2^{1013} \times (2^{1013}-1) \]
Step 4: Divide by the total number of non-empty subsets.
The total number of ways to choose a non-empty $M$ from the $2026$ elements is $2^{2026} - 1$. So:
\[ P(\text{product even}) = \frac{2^{1013}(2^{1013}-1)}{2^{2026}-1} \]
Step 5: Sanity check.
This matches exactly what complementary counting would give, which confirms the direct constructive count is consistent. The key idea is that fixing "at least one even number chosen" while letting the odd numbers be completely free avoids double counting or missing any case.
\[ \boxed{\dfrac{2^{1013}(2^{1013}-1)}{2^{2026}-1}} \]