The response of a linear time invariant system can be split up into a steady state and transient response. For a stable system (all poles of the system have negative real parts) the transient response will go to zero when the time goes to infinity and thus the response would only contain the steady state response. If you would apply some periodic input signal to the system the steady state response would also contain the frequencies present in the input, however for a step response, of step size 1, the steady state would just be a constant defined by the gain of the transfer function of the system at $s=0$.
The steady state response in the given graph seems to tends to 10, so $\left|G(0)\right|=10$ indeed yields $k_m=10$. Your value for the delay $\tau$ is correct as well, assuming the step function jumps to one at $t=0$, be remember to add the units of seconds to this parameter (same units as the time axis). Similar if the y-axis of the figure would have been given an unit, then $k_m$ would require to have this unit as well, but in this case it is dimensionless.
The transient response can be found with the help of the poles of the system, namely for distinct poles the transient response, $y_t(t)$, can be expressed as
$$
y_t(t)=\sum C_i e^{p_it},
$$
where each $p_i$ is a pole of the system and all $C_i$ are constants, which can be solved for with boundary conditions.
I will leave finding the pole of the system up to you, but if you have any questions feel free to ask.