Edit: As Novice has pointed out in a comment below, my answer includes the expressions $(a+b)(c-d)$ and $(a-b)(c+d)$ which, one could argue, shouldn't be included. As the link in Zev Chonoles's answer explains, using two $\pm$ signs in the same expression is ambiguous as it is not clear whether there are two or four possibilities. If you take $(a\pm b)(c\pm d)$ to mean only $(a+b)(c+d)$ and $(a-b)(c-d)$, then we have $$(a\pm b)(c\pm d) = ac \pm ad \pm bc + bd.$$
I don't think there is a concise way to combine them. First note that the expression $(a\pm b)(c\pm d)$ covers four possibilities:
$$(a+b)(c+d) = ac + ad + bc + bd$$
$$(a-b)(c+d) = ac + ad - bc - bd$$
$$(a+b)(c-d) = ac - ad + bc - bd$$
$$(a-b)(c-d) = ac - ad - bc + bd$$
The coefficient of the $ac$ term is always positive ($ac$ may be negative, but the coefficient is always $1$).
The coefficient of the $ad$ term is positive if we choose $+$ in the second bracket (this is because $a$ always has a positive coefficient and now we are choosing $d$ to have a positive coefficient); likewise, the $ad$ term has negative coefficient if we choose $-$ in the second bracket (now we are choosing $d$ to have a negative coefficient).
Similarly, the coefficient of $bc$ is positive if we choose $+$ in the first bracket and negative if we choose $-$.
As you point out, the problem occurs because of the $bd$ term. If the same sign ($+$ or $-$) in both brackets, the coefficient is positive, otherwise it is negative.
As Zev Chonoles has pointed out (and I intended to), you can combine the four expressions using the fact that $a + b = a + (-1)^0b$ and $a - b = a + (-1)^1b$; note that $0$ can be replaced by any even integer, and $1$ can be replaced by any odd integer. If there was only one $\pm$ you have the distribution rules:
$$(a\pm b)(c + d) = a(c+d) \pm b(c+d)$$
$$(a + b)(c \pm d) = (a +b)c \pm (a+b)d$$