Because $ g $ has leading coefficient $ 1 $ and roots $ r^2, $ $ s^2, $ and $ t^2, $ we have \[g(x) = (x-r^2)(x-s^2)(x-t^2)\]for all $ x $. In particular, \[\begin{aligned}-5 = g(-1) &= (-1-r^2)(-1-s^2)(-1-t^2) \\ 5 &= (1+r^2)(1+s^2)(1+t^2).\end{aligned}\]By Vieta's formulas on $ f(x), $ we have $ r+s+t=-a, $ $ rs+st=tr=b, $ and $ rst=1 $. Using this, there are two ways to simplify this sum in terms of $ a $ and $ b $: First option: Expand and repeatedly apply Vieta. We have \[5 = 1 + (r^2+s^2+t^2) + (r^2s^2+s^2t^2+t^2r^2) + r^2s^2t^2.\]We immediately have $ r^2s^2t^2 = (rst)^2 = 1 $. To get $ r^2+s^2+t^2 $ in terms of $ a $ and $ b, $ we write \[r^2+s^2+t^2 = (r+s+t)^2 - 2(rs+st+tr) = a^2 - 2b.\]And to get $ r^2s^2+s^2t^2+t^2r^2 $ in terms of $ a $ and $ b, $ we write \[\begin{aligned} r^2s^2+s^2t^2+t^2r^2 &= (rs+st+tr)^2 - 2(r^2st+rs^2t+rst^2) \\ &= (rs+st+tr)^2 - 2rst(r+s+t)= b^2 + 2a.\end{aligned}\]Thus, \[5= 1 + a^2 - 2b + b^2 + 2a + 1,\]which we can write as \[5 = (a+1)^2 + (b-1)^2.\] Second option: dip into the complex plane. Since $ 1+z^2=(i-z)(-i-z), $ we can rewrite the equation as \[5 = (i-r)(-i-r)(i-s)(-i-s)(i-t)(-i-t).\]Now, for all $ x, $ we have \[f(x) = (x-r)(x-s)(x-t),\]so in particular, $ f(i) = (i-r)(i-s)(i-t) $ and $ f(-i) = (-i-r)(-i-s)(-i-t) $. Thus, \[5 = f(i) f(-i).\]We have $ f(x) = x^3 + ax^2 + bx - 1, $ so \[\begin{aligned} 5 &= (i^3 + ai^2 + bi - 1)((-i)^3 + a(-i)^2 + b(-i) - 1)\\ & =(-(a+1)+ (b-1)i)(-(a+1)- (b-1)i), \end{aligned}\]which simplifies to \[5 = (a+1)^2 + (b-1)^2.\] In either case, the equation we get describes the circle in the $ ab- $plane with center $ (-1, 1) $ and radius $ \sqrt5 $. It follows that the greatest possible value for $ b $ is $ \boxed{1+\sqrt5} $.