For $\:\boldsymbol{\xi}\in \mathcal{H}_{1}\,,\boldsymbol{\eta}\in \mathcal{H}_{2} \:$ and $\:\mathrm{I_{1}}\equiv\text{identity in }\mathcal{H}_{1}\,,\mathrm{I_{2}}\equiv\text{identity in }\mathcal{H}_{2} \:$
\begin{align}
&\left(|a\rangle\langle a|\otimes|b\rangle\langle b|\right)\left(|\boldsymbol{\xi}\rangle\otimes|\boldsymbol{\eta}\rangle\right) =\left(|a\rangle\langle a|\boldsymbol{\xi}\rangle\right)\otimes\left(|b\rangle\langle b|\boldsymbol{\eta}\rangle\right)=
\nonumber\\
& \left(\langle a|\boldsymbol{\xi}\rangle \mathrm{I_{1}}|a\rangle\right)\otimes\left(\langle b|\boldsymbol{\eta}\rangle\mathrm{I_{2}}|b\rangle\right)=
\left[\left(\langle a|\boldsymbol{\xi}\rangle \mathrm{I_{1}}\right)\otimes\left(\langle b|\boldsymbol{\eta}\rangle\mathrm{I_{2}}\right)\right]\left(|a\rangle\otimes|b\rangle\right)=
\tag{01}\\
& \underbrace{\left[\left(\langle a|\mathrm{I_{1}}\right)\otimes\left(\langle b|\mathrm{I_{2}}\right)\right]\left(|\boldsymbol{\xi}\rangle\otimes|\boldsymbol{\eta}\rangle\right)}_{\textrm{scalar quantity}}\left(|a\rangle\otimes|b\rangle\right)=\left(|a\rangle\otimes|b\rangle\right)\underbrace{\left(\langle a|\otimes\langle b|\right)\left(|\boldsymbol{\xi}\rangle\otimes|\boldsymbol{\eta}\rangle\right)}_{\textrm{scalar quantity}}
\nonumber
\end{align}
so
\begin{equation}
\left(|a\rangle\langle a|\otimes|b\rangle\langle b|\right)= \left(|a\rangle\otimes|b\rangle\right)\left(\langle a|\otimes\langle b|\right)
\tag{02}
\end{equation}
Note that a scalar multiple of a vector $\:\mathbf{x}\:$ in a linear space, for example $\:\mathbf{y}=z\cdot\mathbf{x}\:$ where $\:\mathbf{x},\mathbf{y} \in \mathbb{C}^{n} \:$ and $\:z \in \mathbb{C}\:$, could be expressed as
\begin{equation}
\mathbf{y}=\left(z\,\mathrm{I}_{n}\right)\mathbf{x}
\tag{03}
\end{equation}
where $\:\mathrm{I}_{n}\:$ the identity in the linear space, here in $\:\mathbb{C}^{n}$.
Now, the rhs in the 1rst line of (01), that is the term
\begin{equation}
\left(|a\rangle\langle a|\boldsymbol{\xi}\rangle\right)\otimes\left(|b\rangle\langle b|\boldsymbol{\eta}\rangle\right)
\tag{04}
\end{equation}
is essentially a scalar multiple of $\:\left(|a\rangle\otimes|b\rangle\right)\:$
\begin{equation}
\left(|a\rangle\langle a|\boldsymbol{\xi}\rangle\right)\otimes\left(|b\rangle\langle b|\boldsymbol{\eta}\rangle\right)=\underbrace{\langle a|\boldsymbol{\xi}\rangle \langle b|\boldsymbol{\eta}\rangle}_{\textrm{scalar quantity}}\left(|a\rangle\otimes|b\rangle\right)
\tag{05}
\end{equation}
and according to (03) could be expressed as
\begin{equation}
\left(|a\rangle\langle a|\boldsymbol{\xi}\rangle\right)\otimes\left(|b\rangle\langle b|\boldsymbol{\eta}\rangle\right)=\underbrace{\langle a|\boldsymbol{\xi}\rangle \langle b|\boldsymbol{\eta}\rangle}_{\textrm{scalar quantity}}\,\mathrm{I}_{12} \left(|a\rangle\otimes|b\rangle\right)
\tag{06}
\end{equation}
where $\:\mathrm{I}_{12}\:$ the identity in the product space $\:\mathcal{H}_{12}$. But
\begin{equation}
\mathrm{I}_{12}=\mathrm{I}_{1}\otimes\mathrm{I}_{2}
\tag{07}
\end{equation}
This explains the rhs of the 2nd line of (01).(1)
Example
Let $\:\mathcal{H}_{1}\equiv \mathbb{R}^{3} \:$,$\:\mathcal{H}_{2}\equiv \mathbb{R}^{2}$. The terms in the identity (02), repeated here for convenience,
\begin{equation}
\bigl(|\mathbf{a}\rangle\langle \mathbf{a}|\bigr)\boldsymbol{\otimes}\bigl(|\mathbf{b}\rangle\langle \mathbf{b}|\bigr)= \bigl(|\mathbf{a}\rangle\boldsymbol{\otimes}|\mathbf{b}\rangle\bigr)\bigl(\langle \mathbf{a}|\boldsymbol{\otimes}\langle \mathbf{b}|\bigr)
\tag{Ex-01}
\end{equation}
will be expressed explicitly by their components and we'll see a verification of this identity.
So
\begin{align}
&|\mathbf{a}\rangle=
\begin{bmatrix}
a_{1}\\
a_{2}\\
a_{3}
\end{bmatrix} \in \mathbb{R}^{3}\equiv \mathcal{H}_{1}
\, ,\quad
|\mathbf{b}\rangle=
\begin{bmatrix}
b_{1}\\
b_{2}
\end{bmatrix} \in \mathbb{R}^{2}\equiv \mathcal{H}_{2}
\quad
\Longrightarrow
\nonumber\\
&|\mathbf{a}\rangle\boldsymbol{\otimes}|\mathbf{b}\rangle=
\begin{bmatrix}
a_{1}\\
a_{2}\\
a_{3}
\end{bmatrix}
\boldsymbol{\otimes}
\begin{bmatrix}
b_{1}\\
b_{2}
\end{bmatrix}
=
\begin{bmatrix}
a_{1}\begin{bmatrix}
b_{1}\\
b_{2}
\end{bmatrix}\\
\\
a_{2}\begin{bmatrix}
b_{1}\\
b_{2}
\end{bmatrix}\\
\\
a_{3}\begin{bmatrix}
b_{1}\\
b_{2}
\end{bmatrix}
\end{bmatrix}
=
\begin{bmatrix}
a_{1}b_{1}\\
a_{1}b_{2}\\
a_{2}b_{1}\\
a_{2}b_{2}\\
a_{3}b_{1}\\
a_{3}b_{2}
\end{bmatrix} \in \underbrace{\mathbb{R}^{3}\boldsymbol{\otimes}\mathbb{R}^{2}=\mathbb{R}^{6}}_{\:\:\:\mathcal{H}_{1}\:\:\boldsymbol{\otimes}\:\:\mathcal{H}_{2}\:\:\:=\:\:\:\mathcal{H}_{12}}
\tag{Ex-02}
\end{align}
and
\begin{align}
&\langle\mathbf{a}|=
\begin{bmatrix}
a_{1} & a_{2} & a_{3}
\end{bmatrix}
\,,\quad
\langle\mathbf{b}|=
\begin{bmatrix}
b_{1} & b_{2}
\end{bmatrix}
\quad
\Longrightarrow
\langle\mathbf{a}|\boldsymbol{\otimes}\langle\mathbf{b}|=
\begin{bmatrix}
a_{1}\begin{bmatrix}
b_{1} & b_{2}
\end{bmatrix} & a_{2}\begin{bmatrix}
b_{1} & b_{2}
\end{bmatrix} & a_{3}\begin{bmatrix}
b_{1} & b_{2}
\end{bmatrix}
\end{bmatrix}
\nonumber\\
&\Longrightarrow\langle\mathbf{a}|\boldsymbol{\otimes}\langle\mathbf{b}|=
\begin{bmatrix}
a_{1}b_{1}&a_{1}b_{2}&a_{2}b_{1}&a_{2}b_{2}&a_{3}b_{1}&a_{3}b_{2}
\end{bmatrix}
\tag{Ex-03}
\end{align}
Then the transformation in the rhs of (Ex-01) is represented by
\begin{equation}
\bigl(|\mathbf{a}\rangle\boldsymbol{\otimes}|\mathbf{b}\rangle \bigr)\bigl( \langle\mathbf{a}|\boldsymbol{\otimes}\langle\mathbf{b}|\bigr) =
\begin{bmatrix}
a_{1}b_{1}\\
a_{1}b_{2}\\
a_{2}b_{1}\\
a_{2}b_{2}\\
a_{3}b_{1}\\
a_{3}b_{2}
\end{bmatrix}
\begin{bmatrix}
a_{1}b_{1}&a_{1}b_{2}&a_{2}b_{1}&a_{2}b_{2}&a_{3}b_{1}&a_{3}b_{2}
\end{bmatrix}
\tag{Ex-04}
\end{equation}
that is by the $\:6 \times 6\:$ matrix
\begin{equation}
\bigl(|\mathbf{a}\rangle\boldsymbol{\otimes}|\mathbf{b}\rangle \bigr)\bigl( \langle\mathbf{a}|\boldsymbol{\otimes}\langle\mathbf{b}|\bigr) =
\begin{bmatrix}
a_{1}b_{1}a_{1}b_{1} & a_{1}b_{1}a_{1}b_{2} & a_{1}b_{1}a_{2}b_{1} & a_{1}b_{1}a_{2}b_{2} & a_{1}b_{1}a_{3}b_{1}& a_{1}b_{1}a_{3}b_{2}\\
a_{1}b_{2}a_{1}b_{1} & a_{1}b_{2}a_{1}b_{2} & a_{1}b_{2}a_{2}b_{1} & a_{1}b_{2}a_{2}b_{2} & a_{1}b_{2}a_{3}b_{1}& a_{1}b_{2}a_{3}b_{2}\\
a_{2}b_{1}a_{1}b_{1} & a_{2}b_{1}a_{1}b_{2} & a_{2}b_{1}a_{2}b_{1} & a_{2}b_{1}a_{2}b_{2} & a_{2}b_{1}a_{3}b_{1}& a_{2}b_{1}a_{3}b_{2}\\
a_{2}b_{2}a_{1}b_{1} & a_{2}b_{2}a_{1}b_{2} & a_{2}b_{2}a_{2}b_{1} & a_{2}b_{2}a_{2}b_{2} & a_{2}b_{2}a_{3}b_{1}& a_{2}b_{2}a_{3}b_{2}\\
a_{3}b_{1}a_{1}b_{1} & a_{3}b_{1}a_{1}b_{2} & a_{3}b_{1}a_{2}b_{1} & a_{3}b_{1}a_{2}b_{2} & a_{3}b_{1}a_{3}b_{1}& a_{3}b_{1}a_{3}b_{2}\\
a_{3}b_{2}a_{1}b_{1} & a_{3}b_{2}a_{1}b_{2} & a_{3}b_{2}a_{2}b_{1} & a_{3}b_{2}a_{2}b_{2} & a_{3}b_{2}a_{3}b_{1}& a_{3}b_{2}a_{3}b_{2}
\end{bmatrix}
\tag{Ex-05}
\end{equation}
Now,
\begin{equation}
|\mathbf{a}\rangle\langle\mathbf{a}|=\mathbf{a}\mathbf{a}^{\boldsymbol{\top}}=
\begin{bmatrix}
a_{1}\\
a_{2}\\
a_{3}
\end{bmatrix}
\begin{bmatrix}
a_{1} & a_{2} & a_{3}
\end{bmatrix}
=
\begin{bmatrix}
a_{1}a_{1} & a_{1}a_{2} & a_{1}a_{3}\\
a_{2}a_{1} & a_{2}a_{2} & a_{2}a_{3}\\
a_{3}a_{1} & a_{3}a_{2} & a_{3}a_{3}
\end{bmatrix}
\tag{Ex-06}
\end{equation}
and
\begin{equation}
|\mathbf{b}\rangle\langle\mathbf{b}|=\mathbf{b}\mathbf{b}^{\boldsymbol{\top}}=
\begin{bmatrix}
b_{1}\\
b_{2}
\end{bmatrix}
\begin{bmatrix}
b_{1} & b_{2}
\end{bmatrix}
=
\begin{bmatrix}
b_{1}b_{1} & b_{1}b_{2}\\
b_{2}b_{1} & b_{2}b_{2}
\end{bmatrix}
\tag{Ex-07}
\end{equation}
so
\begin{align}
\bigl(|\mathbf{a}\rangle\langle \mathbf{a}|\bigr)\boldsymbol{\otimes}\bigl(|\mathbf{b}\rangle\langle \mathbf{b}|\bigr)&=
\begin{bmatrix}
a_{1}a_{1} & a_{1}a_{2} & a_{1}a_{3}\\
a_{2}a_{1} & a_{2}a_{2} & a_{2}a_{3}\\
a_{3}a_{1} & a_{3}a_{2} & a_{3}a_{3}
\end{bmatrix}
\boldsymbol{\otimes}
\begin{bmatrix}
b_{1}b_{1} & b_{1}b_{2}\\
b_{2}b_{1} & b_{2}b_{2}
\end{bmatrix}
\nonumber\\
&=
\begin{bmatrix}
a_{1}a_{1}\begin{bmatrix}
b_{1}b_{1} & b_{1}b_{2}\\
b_{2}b_{1} & b_{2}b_{2}
\end{bmatrix} & a_{1}a_{2}\begin{bmatrix}
b_{1}b_{1} & b_{1}b_{2}\\
b_{2}b_{1} & b_{2}b_{2}
\end{bmatrix} & a_{1}a_{3}\begin{bmatrix}
b_{1}b_{1} & b_{1}b_{2}\\
b_{2}b_{1} & b_{2}b_{2}
\end{bmatrix}\\
a_{2}a_{1}\begin{bmatrix}
b_{1}b_{1} & b_{1}b_{2}\\
b_{2}b_{1} & b_{2}b_{2}
\end{bmatrix} & a_{2}a_{2}\begin{bmatrix}
b_{1}b_{1} & b_{1}b_{2}\\
b_{2}b_{1} & b_{2}b_{2}
\end{bmatrix} & a_{2}a_{3}\begin{bmatrix}
b_{1}b_{1} & b_{1}b_{2}\\
b_{2}b_{1} & b_{2}b_{2}
\end{bmatrix}\\
a_{3}a_{1}\begin{bmatrix}
b_{1}b_{1} & b_{1}b_{2}\\
b_{2}b_{1} & b_{2}b_{2}
\end{bmatrix} & a_{3}a_{2}\begin{bmatrix}
b_{1}b_{1} & b_{1}b_{2}\\
b_{2}b_{1} & b_{2}b_{2}
\end{bmatrix} & a_{3}a_{3}\begin{bmatrix}
b_{1}b_{1} & b_{1}b_{2}\\
b_{2}b_{1} & b_{2}b_{2}
\end{bmatrix}
\end{bmatrix}
\nonumber\\
&
\tag{Ex-08}
\end{align}
that is
\begin{equation}
\bigl(|\mathbf{a}\rangle\langle \mathbf{a}|\bigr)\boldsymbol{\otimes}\bigl(|\mathbf{b}\rangle\langle \mathbf{b}|\bigr) =
\begin{bmatrix}
a_{1}a_{1}b_{1}b_{1} & a_{1}a_{1}b_{1}b_{2} & a_{1}a_{2}b_{1}b_{1} & a_{1}a_{2}b_{1}b_{2} & a_{1}a_{3}b_{1}b_{1}& a_{1}a_{3}b_{1}b_{2}\\
a_{1}a_{1}b_{2}b_{1} & a_{1}a_{1}b_{2}b_{2} & a_{1}a_{2}b_{2}b_{1} & a_{1}a_{2}b_{2}b_{2} & a_{1}a_{3}b_{2}b_{1}& a_{1}a_{3}b_{2}b_{2}\\
a_{2}a_{1}b_{1}b_{1} & a_{2}a_{1}b_{1}b_{2} & a_{2}a_{2}b_{1}b_{1} & a_{2}a_{2}b_{1}b_{2} & a_{2}a_{3}b_{1}b_{1}& a_{2}a_{3}b_{1}b_{2}\\
a_{2}a_{1}b_{2}b_{1} & a_{2}a_{1}b_{2}b_{2} & a_{2}a_{2}b_{2}b_{1} & a_{2}a_{2}b_{2}b_{2} & a_{2}a_{3}b_{2}b_{1}& a_{2}a_{3}b_{2}b_{2}\\
a_{3}a_{1}b_{1}b_{1} & a_{3}a_{1}b_{1}b_{2} & a_{3}a_{2}b_{1}b_{1} & a_{3}a_{2}b_{1}b_{2} & a_{3}a_{3}b_{1}b_{1}& a_{3}a_{3}b_{1}b_{2}\\
a_{3}a_{1}b_{2}b_{1} & a_{3}a_{1}b_{2}b_{2} & a_{3}a_{2}b_{2}b_{1} & a_{3}a_{2}b_{2}b_{2} & a_{3}a_{3}b_{2}b_{1}& a_{3}a_{3}b_{2}b_{2}
\end{bmatrix}
\tag{Ex-09}
\end{equation}
The matrices in the right hand sides of equations (Ex-05) and (Ex-09) are equal, so the left hand sides, verifying the validity of the identity (Ex-01).
(1)
I suggest to take a look in my 2nd and 3rd answers therein : Total spin of two spin-1/2 particles about product states, product spaces and product transformations which, although used for spin, give informations and details in general.
