jmc

The digits $2$, $4$, $5$, and $6$ cannot be the units digit of any two-digit prime, so these four digits must be the tens digits, and $1$, $3$, $7$, and $9$ are the units digits. The sum is thus $$10(2+4+5+6)+(1+3+7+9)=\boxed{190}.$$(One set that satisfies the conditions is $\{23,47,59,61\}$.)