Overshoot more than 100%

Write the TF as \$ \small G(s)=\large\frac{ab}{(s+a)(s+b)}\$, where \$a\$ and \$b\$ are real and distinct.

This factorises to: \$\small G(s)=\large\frac{ab}{(b-a)} \large(\frac{1}{s+a}-\frac{1}{s+b}\large)\$

From which the inverse LT, i.e. the unit impulse response, is:

\$g(t)= \large\frac{ab}{(b-a)} \normalsize (e^{-at}-e^{-bt})\$

Differentiating and setting to zero determines the time to peak value, \$t_P\$:

\$t_P =\large\frac{ln(\frac{b}{a})}{(b-a)}\$, with corresponding peak value: \$g(t_P)=be^{-b t_P}\$

Hence, measuring \$t_P\$ and \$g(t_P)\$ will yield the value of \$b\$; and substituting back into the expression for \$t_P\$ will yield \$a\$

Note that the unit impulse response can be obtained by differentiating the unit step response, but the differentiator must be carefully designed to avoid introducing excessive noise.

Typical overdamped step and impulse responses are shown below.

I got the position of the pole to be \$8.873\$ and the gain to be \$8.873\$.

First,

yields \$z = 0.8873\$. Next plug in \$z\$ into the transfer function above and you get \$-0.1127\$. The gain is then \$\frac{1}{0.1127} = 8.873\$.

Step response is as follows

I also had issues with the overshoot (about \$33 \%\$) but at least it meets the pole-zero relationship. I'm guessing it's an issue with the two poles at \$0\$.