Electric fields obey the superposition principle. So let's call the two fields $\mathbf{E}_{1}\left( \omega_{1} \right)$ and $\mathbf{E}_{2}\left( \omega_{2} \right)$. Then the total field, $\mathbf{E}_{t}$, is given by:
$$
\mathbf{E}_{t} = \mathbf{E}_{1} + \mathbf{E}_{2} \tag{1}
$$
Note that the field intensity or energy density is given by:
$$
W = \frac{ \varepsilon_{o} \ \left( \mathbf{E} \cdot \mathbf{E} \right) }{ 2 } \tag{2}
$$
where $\varepsilon_{o}$ is the permittivity of free space. You can see that $W_{t} \neq W_{1} + W_{2}$ by noticing that:
$$
\begin{align}
\left( \mathbf{E}_{t} \cdot \mathbf{E}_{t} \right) & = \left( \mathbf{E}_{1} + \mathbf{E}_{2} \right) \cdot \left( \mathbf{E}_{1} + \mathbf{E}_{2} \right) \tag{3a} \\
& = \left( \mathbf{E}_{1} \cdot \mathbf{E}_{1} \right) + \left( \mathbf{E}_{2} \cdot \mathbf{E}_{2} \right) + 2 \left( \mathbf{E}_{1} \cdot \mathbf{E}_{2} \right) \tag{3b} \\
& \neq \left( \mathbf{E}_{1} \cdot \mathbf{E}_{1} \right) + \left( \mathbf{E}_{2} \cdot \mathbf{E}_{2} \right) \tag{3c}
\end{align}
$$
Is it possible to calculate the intensity of the total electric field?
Yes, add the two input fields together then calculate the intensity using Equation 2 above.
I would like to calculate the specific absorption rate (SAR), but for that I need the magnitude of the total electric field.
Calculate $\mathbf{E}_{t}$ using Equation 1 above and then find its magnitude, which is the total electric field magnitude.