Clearly, the number of total possibilities is $10\times 10=100$.

There are multiple ways to count the number of qualified cases:

List all the cases depending on whether each draw is $7$

Therefore, the total count is $1 + 9 + 9 = 19$ and the probability is $\boxed{\frac{19}{100}}$.

A little improvement over the previous method

We note that if the first card is $7$ then we do not need to care about the second card. Therefore, the total count can also be calculated as $$1\times 10 + 9\times 1 = 19$$

Count the opposite

If and only if neither card is $7$ will make their product not a multiple of $7$. There are $9\times 9=81$ such cases. Therefore, the qualified cases count is $100-81=19$.

The probability way

The event that the first card drawn is $7$ and the second card drawn is $7$ are indpendent events with probability of $\frac{1}{10}$. Therefore the probability that at least one of them occurs is $$\frac{1}{10} +\frac{1}{10} -\frac{1}{10}\times\frac{1}{10}=\frac{19}{1000}$$