I have skipped some basic math steps to try and make my answer as short as I can. If you do not understand, comment down below and I will make myself a bit more clear.
I do not mean to steal the answer of @QwirkyQwilfish as he/she was so close, but I believe the operator is
If you have two numbers each with $m$ and $n$ digits respectively, with $x$ letters and $y$ letters respectively, then the operator is equal to
$n\times (m+x-1)$ if $m>n$
Unknown if $m=n$
$m\times (n+y-1)$ if $m<n$
Supporting Examples:
$1$ 🎫 $1 = \underbrace{\text{one}}_{3}\times \underbrace{\text{one}}_{3} = 3\times 3 = 9$.
$1$ 🎫 $2=\underbrace{\text{one}}_{3}\times \underbrace{\text{two}}_{3} = 3\times 3 = 9$.
$3$ 🎫 $4 = \underbrace{\text{three}}_{5}\times \underbrace{\text{four}}_{4}=5\times 4 = 20 \to\require{cancel}{\cancel2}0=0$. $\qquad m=5, n=4$.
$2$ 🎫 $6 = \underbrace{\text{two}}_{3}\times \underbrace{\text{six}}_{3}=3\times 3 = 9$.
$7$ 🎫 $10 = \underbrace{\text{seven}}_{5}\times \underbrace{\text{ten}}_{3} = 5\times 3 = 15 \to \cancel{1}5=5$. $\qquad m=5, n=3$.
$10$ is two digits, but we disregard that because it does not have the greatest letter-amount.
Now, regarding the second half, it is a bit different because the numbers have two digits instead of $1$. But we can still use the same rule.
Supporting Examples
$15$ 🎫 $12=\underbrace{\text{fifteen}}_{7}\times \underbrace{\text{twelve}}_{6}=(7+1)\times 6 = 8\times 6 = 48\to \cancel{4}8 = 8$. $\; m=7, n=6$.
$34$ 🎫 $40=\underbrace{\text{thirty-four}}_{11}\times \underbrace{\text{forty}}_{5} = (11+1)\times 5 = 12\times 5 = 60\to \cancel{6}0=0.$
As you can see above, the dash/hyphen "$-$" counts as a letter. Moving on,
$80$ 🎫 $203= \underbrace{\text{eighty}}_{6}\times \underbrace{\text{two-hundred-and-three}}_{18}=(18+2)\times 6 = 20\times 6=120\to \cancel{12}0=0$.
In this example, $203$ has three digits, so we add $2$ to the greatest number, and we disregard $80$ with two digits because its letter-amount is less than the letter-amount of $203$, i.e. we do not add $1$ to $6$ because it is not the greatest number. Overall, the general formula is stated in the first sandbox above.
The last answer is:
$$\begin{align}491 \, 🎫 \, 705&=\underbrace{\text{four-hundred-and-ninety-one}}_{27}\times \underbrace{\text{seven-hundred-and-one}}_{21} \\ \\ &= (27+3-1)\times 21\qquad \binom{\text{since } 27>21 \text{ and}}{491 \text{ has } 3 \text{ digits }} \\ \\ &= (27+2)\times 21 \\ \\ &= 29\times 21 \\ \\ &= 609\to \cancel{60}9=9.\end{align}$$
