Since $f(x) = ax + b$ for all real numbers $x$, then $f(t) = at + b$ for some real number $t$. When $t = bx + a$, we obtain $f(bx + a) = a(bx + a) + b = abx + (a^{2} + b)$. We also know that $f(bx + a) = x$ for all real numbers $x$. This means that $abx + (a^{2} + b) = x$ for all real numbers $x$ and so $(ab - 1)x + (a^{2} + b) = 0$ for all real numbers $x$. For this to be true, it must be the case that $ab = 1$ and $a^{2} + b = 0$. From the second equation $b = -a^{2}$ which gives $a(-a^{2}) = 1$ and so $a^{3} = -1$, which means that $a = -1$. Since $b = -a^{2}$, then $b = -1$ as well, which gives $a+b = -2$.