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The first equality you wrote is not correct, as noted by other users. However, what I think you want to know is why $(I:A)I = (I \otimes I)A$. You can show this just by using dyads properties. $$(B \otimes B) : A =(B_{ij}B_{kl} e_i \otimes e_j \otimes e_k \otimes e_l):(A_{mn} e_m \otimes e_l) = B_{ij}B_{kl} A_{mn} \delta_{km} \delta_{ln} e_i \otimes e_j = B_{...

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