As easy as ABC!

AB x AA = BAAB

(10 A + B) (10 A + A) = 1000 B + 100 A + 10 A + B

(10A + B) 11 A = 1001 B + 110 A

110 A² + 11 AB = 1001 B + 110 A

10 A² + AB = 91 B + 10 A [dividing by 11]

10 A² + AB - 10 A = 91 B

10 (A² - A) = (91 - A) B

B = 10 $\frac{(A^2 - A)}{(91 - A)}$

Trying each A (0-9)

A = 1, B = 0;

A = 7, B = 5;

Since 10 x 11 = 110 $\neq$ a four-digit number

$A=\boxed{7 }$

$B=\boxed{5}$

75 * 77 = 5775.

$\left( 10A+B \right) \cdot \left( 10A+A \right) =B+10A+100A+1000B$

$\left( 10A+B \right) \cdot 11A=11\cdot \left( 91B+10A \right)$

$10{ A }^{ 2 }+AB=91B+10A$

$10{ A }^{ 2 }+AB-10A=91\cdot B$

$A(10A+B-10)=91\cdot B$

And since $A|91\cdot B$ and $A\le 9$ then $A=B$ or $A=7$ . In the first case we get that $11A=101$ which is a contradiction and in the second we get that $B=5$ .

So $A=7$ and $B=5$