$
\def\a{\alpha}\def\d{\delta}\def\e{\varepsilon}
\def\s{\star}\def\o{\otimes}
\def\p{\partial}
\def\A{{\cal A}}\def\B{{\cal B}}\def\C{{\cal C}}
\def\E{{\cal E}}\def\F{{\cal F}}\def\G{{\cal G}}
\def\I{{\mathbb I}}\def\S{{\cal S}}
\def\LR#1{\left(#1\right)}
\def\op#1{\operatorname{#1}}
\def\trace#1{\op{tr}\LR{#1}}
\def\sym#1{\op{sym}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\qif{\quad\iff\quad}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\m#1{\left[\begin{array}{r}#1\end{array}\right]}
\def\c#1{\color{red}{#1}}
\def\CLR#1{\c{\LR{#1}}}
\def\fracLR#1#2{\LR{\frac{#1}{#2}}}
$While common in Elasticity and Continuum Mechanics, the symbol $\o$ is used for the Kronecker product rather than the dyadic product in most other fields.
This is because those other fields rarely require a dyadic product, but the Kronecker product is extremely useful for flattening matrices into vectors $\,($and tensors into matrices, thereby avoiding tensors altogether$)$.
I would also mention that in physics, simple juxtaposition is often used for dyadics, i.e.
$$a\o b\o c \implies abc$$
A better choice is Index Notation, which is utilized in all fields. It's also clear and compact, especially if you omit the superfluous cartesian basis $\c{\rm vectors}$
$$a\o b\o c = a_ib_jc_k\CLR{e_i\o e_j\o e_k} \qiq a_ib_jc_k\\$$
Now to answer your question, there are three different isotropic fourth-order tensors, all of which can be expressed in terms of the Kronecker delta symbol
$$\eqalign{
\d_{jk}\d_{\ell m} = \G_{jk\ell m} = \F_{j\ell mk} = \E_{j\ell km}
}$$
Note that $\G$ is simply the dyadic product of two identity matrices,
while $\E$ and $\F$ are obtained by permuting the indices. Every component of these tensors is $\,\in\{0,{\tt1}\}$
Just as the single dot product is the sum over a single adjacent index, the double-dot product is the sum over an ordered pair of adjacent indices, e.g.
$$\eqalign{
\def\sk{\sum_{k=1}^m}
\def\sl{\sum_{\ell=1}^n}
C &= A\cdot B \qiq C_{ij} = \sk A_{i\c{k}}\,B_{\c{k}j} \\
\C &= \A:\B \qiq \C_{ijpq} = \sk\sl \A_{ij\c{k\ell}}\,\B_{\c{k\ell}pq} \\
M &= N:\A \qiq M_{ij} = \sk\sl N_{\c{k\ell}}\,\A_{\c{k\ell}ij} \\
\|A\|_F^2 &= A:A \qiq \|A\|_F^2 = \sk\sl A_{\c{k\ell}}\,A_{\c{k\ell}} \\
}$$
The product of an arbitrary matrix $M$ with the fourth-order isotropic tensors yields
$$\eqalign{
\E:M &= M:\E &= M \\
\F:M &= M:\F &= M^T \\
\G:M &= M:\G &= I\,\trace{M} \\
}$$
Thus $\E$ is sometimes called the identity tensor since it acts as the identity here.
Note that the $\I$ in your post actually equals $\G$, which is one of the isotropic tensors but not the identity tensor (although it does yield a scalar multiple of the identity matrix).
Some authors refer to the following $\S$ tensor as the identity tensor
$$\eqalign{
\S &= \tfrac 12\LR{\E+\F} \\
M:\S &= \S:M = \frac{M+M^T}{2} \;\equiv\; \sym{M} \\
}$$
but as you can see, it acts as an identity only if the matrix $M$ is symmetric.
Another useful property of the isotropic tensors is their ability to rearrange matrix-vector products. For example
$$\eqalign{
P\cdot X\cdot Q &= \LR{P\cdot\E\cdot Q^T}:X \\
P\cdot X\cdot q &= \LR{P\cdot\E\cdot q}:\F:X
&= \LR{P\cdot\G\cdot q}:X \\
p\cdot X\cdot Q &= \LR{p\cdot\E\cdot Q^T}:X
&= X:\LR{p\cdot\G\cdot Q} \\
p\cdot X\cdot q &= \LR{p\cdot\G\cdot q}:X
&= \LR{p\cdot q^T}:X \\
p\cdot x^T\cdot Q &= \LR{p\cdot\G\cdot Q^T}\cdot x \\
P\cdot x\cdot q^T &= \LR{P\cdot\E\cdot q}\cdot x \\
}$$
where $(x,p,q)$ are vectors and $(X,P,Q)$ are matrices.
