For currents of 100mA or so, at 1kHz or less, I don't think you need much more to make that Howland-like current source work well, using almost any op-amp:
simulate this circuit – Schematic created using CircuitLab
Q1 and Q2 are a push-pull buffer to alleviate the op-amp of responsibility for current. It will have some cross-over distortion, but since your inputs are stepped, that shouldn't be a concern.
We need \$\frac{R_3}{R_4}=\frac{R_5}{R_6}\$ to keep common-mode gain near zero. Output current \$I_{OUT}\$ will be:
$$ I_{OUT} = V_{IN}\frac{R_3}{R_4}\cdot \frac{1}{R_2} $$
Using the values shown, this evaluates to:
$$ I_{OUT} = \frac{V_{IN}}{10\Omega} $$
This is an approximation, since some of that current will flow via R5, but as long as you keep R5 >> R2, it's very close. Using the values shown, error should be within 0.5%, which is still smaller than any error you might see due to resistor tolerances.
This design can only apply ±11V across load \$R_L\$, at most, so the largest resistance load that you'll be able to push 100mA through is:
$$ R_{L(MAX,100mA)} = \frac{11V}{100mA} = 110\Omega $$
Also, don't forget that Q1 and Q2 are dissipating a couple of watts, potentially, so they should be power devices (like the TIP models shown), and you may need small heatsinks on them.
There will be some crossover distortion, due to the absence of biasing of Q1 and Q2, but since your inputs are stepped, this shouldn't be a problem. In any case, you can mitigate crossover distortion by using a high output slew rate op-amp, like the TL081 shown. Even a slow device like the LM358 should work here, at 1kHz.
Here is load current for a stepped input \$V_{IN}\$ with values 0V, 1V, -0.75V and +0.5V:
Output current transitions are slowed somewhat by C1 and C2, which try to keep the closed loop stable for badly-behaved loads.