Help with a Closed loop and zero order hold function ZoH * 1/s(s+a) type
I haven't checked your solution in any detail, but my analysis gives the OLTF:

$$\small G_{OL}=\frac{z}{z-1}\: \large \mathcal {Z}\small \left( \frac{a}{s^2(s+a)}\right ) = \frac{z}{z-1}\: \large \mathcal {Z}\small \left( \frac{1}{s^2}  -\frac{1/a}{s}  +\frac{1/a}{s+a}\right ) $$

taking z-transforms:

$$\small G_{OL}(z)= \frac{z-1}{z}\left( \frac{Tz}{(z-1)^2}-\frac{1}{a}\frac{z}{(z-1)}+\frac{1}{a}\frac{z}{(z-e^{-aT})}\right )$$

$$\small \therefore \: G_{OL}(z)= \frac{T}{(z-1)}-\frac{1}{a}+\frac{1}{a}\frac{z-1}{(z-e^{-aT})}$$

and the CLTF, taking \$\small T=1\$:
$$ \small G(z)=\frac{(a+e^{-a}-1)z+(1-e^{-a}-ae^{-a})}{az^2+(e^{-a}-1-ae^{-a})z+(1-e^{-a}) }$$

In regard to your parameter values, \$\small a=T=1\$ gives a critically stable closed loop (pole on the unit circle). However, taking \$\small a=2; T=1\$, gives the CLTF:
$$ \small G(z)=\frac{0.57z+0.29}{z^2-0.57z+0.43}$$
which factorises to:

$$ \small G(z)=\frac{0.57z+0.29}{(z-0.285+ j0.59)(z-0.285- j0.59)}$$
and, with poles inside the unit circle, this CLTF is stable.

To plot the unit step response, we may obtain the difference equation from \$\small G(z)\$, thus:

$$\small y[k]=0.57x[k-1]+0.29x[k-2]+0.57y[k-1]-0.43y[k-2]$$

Excel produces the following step response from the difference equation: