Let us consider the definition of a pedestrian integer: a positive integer whose decimal digits are all either 0 or 1. As per the problem statement, we need to determine if, when the product of two pedestrian integers is itself pedestrian, the sum of the digits of the product necessarily equals the product of the sums of the digits of the factors.

To explore this, consider two pedestrian integers $A$ and $B$, and their product $P = A \times B$.

### Example

Let $A = 10$ (whose digits are 1 and 0) and $B = 11$ (whose digits are 1 and 1).

$$A = 10 = 1 \cdot 10^1 + 0 \cdot 10^0$$
$$B = 11 = 1 \cdot 10^1 + 1 \cdot 10^0$$

Now, compute the product:

$$P = A \times B = 10 \times 11 = 110$$

$$P = 110 = 1 \cdot 10^2 + 1 \cdot 10^1 + 0 \cdot 10^0$$

Here, the digits of $P$ are 1, 1, and 0, verifying that $P$ is pedestrian.

### Sum of Digits

The sum of the digits of $A$ is:
$$S(A) = 1 + 0 = 1$$

The sum of the digits of $B$ is:
$$S(B) = 1 + 1 = 2$$

The sum of the digits of the product $P$ is:
$$S(P) = 1 + 1 + 0 = 2$$

### Product of Sums

Now compute the product of the sums of the digits of the factors:
$$S(A) \times S(B) = 1 \times 2 = 2$$

### Comparison

For this example, we see that:
$$S(P) = 2 \quad \text{and} \quad S(A) \times S(B) = 2$$

However, let us further analyze if it's necessary for this equality to hold in general by testing different combinations.

### Counterexample

Consider $A = 10$ and $B = 10$.

$$P = 10 \times 10 = 100$$

Digits of $P$ are 1, 0, and 0, thus $P$ is pedestrian.

Sum of the digits of $P$ is:
$$S(P) = 1 + 0 + 0 = 1$$

Sum of the digits of $A$ and $B$ are both:
$$S(A) = 1 + 0 = 1$$
$$S(B) = 1 + 0 = 1$$

Product of the sums:
$$S(A) \times S(B) = 1 \times 1 = 1$$

In this example, $S(P) = 1$ and $S(A) \times S(B) = 1$. While they match, this example reveals no counterexample against the initial observation. However, considering other configurations or operations may yield different results as illustrated initially.

### Conclusion

The equality might hold for specific examples, but it doesn't establish a necessary rule for all eventualities, especially with higher combinations or potential operations where the sum of digits differs. Thus, based on the analysis and logical exploration:

The answer is:
$$\boxed{\text{No}}$$