The assumption behind the Saint Venant-Kirchhoff constitutive model is that the energy per unit volume of the material in the reference configuration (i.e. strain energy density) can be adequately approximated as a quadratic function of the Lagrangian Green strain $E$. 469
(The logic is similar to the approximation of a point particle's potential energy being quadratic near a stable equilibrium, but is ultimately guided by experimental validation.)
Therefore, the generic form of the strain energy density in the Saint Venant-Kirchhoff model, in index notation, is of the following form: $$W = \frac{1}{2}C_{ijkl}E_{ij}E_{kl}$$ where $C = C_{ijkl}$ is a fourth-order tensor of elastic constants. Note that, due to the symmetry of the Lagrangian Green strain tensor and the fact that $W$ is a scalar, $C$ also possesses quite a few internal symmetries as well; $C_{ijkl} = C_{jikl} = C_{ijlk} = C_{lkij}$.
If we take the partial derivative of this general expression with respect to the Lagrangian Green strain tensor, we find: $$\frac{\partial W}{\partial E_{mn}} = \frac{1}{2}C_{ijkl}\frac{\partial E_{ij}}{\partial E_{mn}}E_{kl} + \frac{1}{2}C_{ijkl}E_{ij}\frac{\partial E_{kl}}{\partial E_{mn}}$$ This results in a few <PERSON> deltas: $$\frac{\partial W}{\partial E_{mn}} = \frac{1}{2}C_{ijkl}\delta_{im}\delta_{jn}E_{kl} + \frac{1}{2}C_{ijkl}E_{ij}\delta_{km}\delta_{ln}$$ $$\frac{\partial W}{\partial E_{mn}} = \frac{1}{2}C_{mnkl}E_{kl} + \frac{1}{2}C_{ijmn}E_{ij}$$ Since the Lagrangian green strain tensor is symmetric ($E_{ij} = E_{ji}$) and $C$ possesses the symmetries listed above, the two terms combine to make: $$\frac{\partial W}{\partial E_{mn}} = S_{mn} = C_{mnkl}E_{kl}$$ Or, in index-free notation, $$S = C:E$$
Now suppose that you assume the material is also isotropic; that is, $C$ is an isotropic tensor. As it turns out, the most generic form for a fourth-order isotropic tensor $C$ would be: $$C_{ijkl} = \lambda\delta_{ij}\delta_{kl} + \nu_1 \delta_{ik}\delta_{jl} + \nu_2 \delta_{il}\delta_{jk}$$ where $\lambda$, $\nu_1$ and $\nu_2$ are unknown parameters.
Inputting this into the expression for $S$ above, noting again that $E_{ij} = E_{ji}$, and defining $\mu = \nu_1 + \nu_2$, we find that:
$$S_{ij} = \lambda (E_{kk}) \delta_{ij} + \nu_1 E_{ij} + \nu_2 E_{ji}$$ $$S_{ij} = \lambda (E_{kk}) \delta_{ij} + 2 \mu E_{ij}$$
Or, in index-free notation,
$$S = \lambda\ \text{tr}E\ I + 2 \mu E$$
Instead of making the assertion that $C$ is isotropic after deriving a general expression for $S$, you can simply take the statement that the Saint Venant-Kirchhoff model for the strain energy density is accurate based on empirical modeling and then derive an expression for $S$ from that. 649