Let $a = \overline{x7}$ and $b = \overline{yd}$, where $d$ is the final digit of $b$ and $x, y$ are the numbers $a$ and $b$ without their final digits. Then $a = 10x + 7$ and $b = 10y + d$. From the given information, we compile the system of equations
$$\begin{cases} x + 10y + d = 2022, \\ y + 10x + 7 = 5000. \end{cases} \qquad (1)$$
Subtracting the first equation from the second one, we get the equation
$$9(x - y) - d = 2971. \qquad (2)$$
As $x, y$ are integers, $9(x - y)$ is divisible by 9. As the number 2971 gives a remainder of 1 upon division by 9, the number $d$ must give a remainder of 8; the only option for this is $d = 8$. Substituting this into (2), we get $x - y = 331$, which we can substitute into either equation of (1) to get $x = 484$ and $y = 153$. Therefore $a + b = 4847 + 1538 = 6385$.