You probably know that while the Cauchy stresses are objective, its stress rate (material derivative) is not. If $Q(t)$ is an orthogonal tensor representing a change of frame, the stress is the new frame is
$$
T^* = Q T Q^T
$$
However, if you take material derivatives on both sides, you have,
$$
\dot{T^*} = \dot{Q} T Q^T + Q \dot{T} Q^T + Q T \dot{Q}^T \quad \quad (1)
$$
Terms 1 and 3 above are "extra" -> they imply that naively taking the material derivative of the stress $T$ won't get you an objective rate.
A similar problem arises because the "spin" part of the velocity gradient tensor $L$, (I'll call it $W$ rather than $\Omega$) is not objective.
i.e., $L = W + D$, but while $D$ is objective, $W$ is not. Starting from first principles, it is not hard to show that (I'll add this bit if you want)
$$
W^* = Q W Q^T + \dot{Q} Q^T \quad \quad (2)
$$
Also, because $Q$ is orthogonal ($Q^T=Q^{-1}$) you have the lemma
$$
\dot{Q} Q^T = -Q \dot{Q}^T
$$
We can actually use this second problem to construct a solution to the first. We'll eliminate $\dot{Q}, \dot{Q}^T$ from (1) in turn.
Right multiply (2) by Q and rearrange; you have
$$
\dot{Q} = W^* Q - Q W
$$
Transpose both sides and you have an expression for $\dot{Q}^T$.
$$
\dot{Q}^T = -Q^T W^* + W Q^T
$$
The last result uses the fact that the spin tensor $W$ is skew-symmetric.
Let's substitute these expressions for $\dot{Q},\dot{Q}^T$ back in (1).
$$
\dot{T^*} = (W^* Q - Q W) T Q^T + Q \dot{T} Q^T + Q T (-Q^T W^* + W Q^T) \\
\quad = W^* Q T Q^T - Q W T Q^T + Q \dot{T} Q^T - Q T Q^T W^* + Q T W Q^T) \\
\quad = W^* T^* - Q W T Q^T + Q \dot{T} Q^T - T^* W^* + Q T W Q^T \\
\Rightarrow \quad \text {(on rearranging)} \\
\quad \dot{T^*} + T^* W^* - W^* T^* = Q (\dot{T} + T W - W T) Q^T \\
$$
Notice that the right had side has the form $Q R Q^T$, and the left side is precisely the bracketed quantity in the starred frame of reference. In other words, we have constructed a rate $J = \dot{T} + T W - W T$ such that
$\dot{J^*} = Q J Q^T $, which is objective by definition. You can now easily specialize your equations to elasticity.
The important thing to remember is that the Jaumann rate is merely one of an infinite number of objective stress rates. In general, Lie derivatives of objective tensor fields are objective.
It seems that you confused the Jaumann derivative $\overset{o}{{S}}$ (in your notation $\overset{\bigtriangleup}{{S}}$) with the time derivative ${\dot{S}}$
$$\frac{dS}{dt} = {\dot{S}} = \overset{o}{{S}} -{S} \cdot {w} +{w} \cdot {S}$$
See how it is derived in "http://www.continuummechanics.org/cm/corotationalderivative.html". Using the argument that $w^T = -w$ we get
$$\frac{dS}{dt} = {\dot{S}} = \overset{o}{{S}} +{S} \cdot {w^T} +{w} \cdot {S}$$
Moreover, when you "clarified" where does the deviatoric strain come from it is not quite clear to me all the steps you followed. Because actually is not the time derivative that equals the term $\left({\dot{{\epsilon}}}^{ij} - \frac{1}{3}{\delta}^{ij}{\dot{{\epsilon}}}^{ij} \right)$ but the Jaumann derivative. For this you can take a look at Hookes law and objective stress rates
More concretely, the derivation of the formula for the Jaumann stress
$$\overset{o}{{S}^{ij}} = 2\mu\left[{\overset{o}{\epsilon}^{ij}} - \frac{1}{3}\delta^{ij}{\overset{o}{\epsilon}^{ij}}\right]$$
is derived in Hooke's law wikipedia and in "An Introduction to Continuous Mechanics, Klaus Hackl, Mehdi Goodarzi". However, here I show a summary of the derivation of the Jaumann stress.
The definition of deviatoric stress is just
$$\sigma = \begin{pmatrix}
\frac{\sigma_{11}}{3} & 0 & 0\\
0 & \frac{\sigma_{22}}{3} & 0\\
0 & 0 & \frac{\sigma_{33}}{3}\\
\end{pmatrix} + \begin{pmatrix}
\sigma_{11} - \frac{\sigma_{11}}{3} & \sigma_{12} & \sigma_{13}\\
\sigma_{21} & \sigma_{22} - \frac{\sigma_{22}}{3} & \sigma_{23}\\
\sigma_{31} & \sigma_{32} & \sigma_{33}- \frac{\sigma_{33}}{3}\\
\end{pmatrix}$$
the second term of the previous equation. If we apply Hooke's law to it (which relates stress $\sigma$ and strain $\varepsilon$ linearly)
$${{S}} = 2\mu\left({{\epsilon} - \frac{1}{3}\mathrm{tr}(\varepsilon)}\right)$$
With the Einstein notation
$${{S}}^{ij} = 2\mu\left[{{\epsilon}}^{ij} - \frac{1}{3}\delta^{ij}\epsilon^k{}_k\right]$$
where $\epsilon$ is the so called strain tensor.
Finally, we can compute the Jaumann derivative to get what we wanted.
Best Answer
First, I'm going to define some notation. I prefer the direct notation of modern continuum mechanics as presented in this book by Gurtin et al., so I won't be carrying around the indices that you use. Lastly, the following holds for small deformation, rate-independent plasticity with isotropic hardening.
Next, we write out the Kuhn-Tucker consistency condition at yield (which occurs at $f=0$):
This may seem like a strong statement, and it is, but if you think about all of the possible types of loading from a point on the "yield surface" (defined as all stress states that give $f=0$), you will find this to be true.
Now, we apply this condition to find $\dot{\bar{\epsilon}}^p$. First, compute $\dot{f}$.
\begin{align} \dot{f} &= \sqrt{\frac{3}{2}} \dot{\bar{\sigma}} - \dot{Y}\\ &= \sqrt{\frac{3}{2}} \frac{\dot{\mathbf{T}}_0 : \mathbf{T}_0}{|\mathbf{T}_0|} - H(\bar{\epsilon}^p)\dot{\bar{\epsilon}}^p\\ &= \sqrt{\frac{3}{2}} \dot{\mathbf{T}}_0 : \mathbf{N}^p - H(\bar{\epsilon}^p)\dot{\bar{\epsilon}}^p \\ &= \left(\sqrt{\frac{3}{2}}\right)(2G)[\dot{\mathbf{E}} - \dot{\mathbf{E}}^p] : \mathbf{N}^p - H(\bar{\epsilon}^p)\dot{\bar{\epsilon}}^p \\ &= \left(\sqrt{\frac{3}{2}}\right)(2G)\dot{\mathbf{E}} : \mathbf{N}^p - (3G + H)\dot{\bar{\epsilon}}^p \end{align}
Now, we have to consider three cases: 1) Elastic unloading, 2) Neutral Loading, 3) Plastic Loading
$$ \dot{\bar{\epsilon}}^p = \frac{\sqrt{3/2} (2G) (\dot{\mathbf{E}}:\mathbf{N}^p)}{3G + H(\bar{\epsilon}^p)} $$
I glossed over some of the more tedius algebra and buried the assumption that $(3G + H)>0$, which is true for every case that I have encountered. You can go on from here to compute the so-called "elasto-plastic" modulus, which is necessary to do implicit integration in a finite element method.