Multiplexer diagram of \$C = A \cdot B\$
You can do it by putting 0 in the first input, \$A\$ in the second input, and \$B\$ in the select bit. Hence
$$C = \bar{B} 0 + BA = AB.$$

If you set the first input to \$A\$, the second input to \$B\$, and the select bit to \$\bar{A}\$ you get
$$A + \bar{A}B = A(B + 1) + \bar{A} B = A + AB +\bar{A} B = A + (A + \bar{A}) B = A + B.$$

It should be noted that if you set the select bit to \$A\$, as you have done, you get
$$\bar{A}A + AB = AB,$$
not \$A + B\$.