For clarification, following the question, let us emphasize some notations first.
Let $M_n$ be the collection of $n\times n$ complex valued matrices, let $H_n$ be the collection of $n\times n$ complex valued Hermitian matrices, and given $X,Y\in H_n$, denote $X\ge 0\Leftrightarrow$ $X$ is positive semidefinite, and $X\ge Y\Leftrightarrow X-Y\ge 0$. Given $m,n$, for $X\in M_{mn}$, write $X=(X_{ij})_{n\times n}$, where $X_{ij}\in M_m$. Define
$${\rm tr}_1 (X)=\sum_{i=1}^n X_{ii}\in M_m,\quad {\rm tr}_2 (X)=({\rm tr}(X_{ij}))_{n\times n}\in M_n\tag{1}$$
and
$$\mathcal P(X)={\rm tr}_2(X)\otimes {\rm tr}_1(X)\in M_{mn}.\tag{2}$$
Note that the definitions $(1)$ and $(2)$ for $X\in M_{mn}(X)$ not only depend on the the product $mn$, but also depend on $m$ and $n$ themselves. Since there is no much ambiguity in the following, for simplicity, we will not specify the dependence on $m,n$.
Claim: Given $X\in H_{mn}$, $$X\ge 0\Rightarrow {\rm Im}(\mathcal P(X))\supset {\rm Im}(X).\tag{3}$$
The proof is based on the following facts. Since the argument is very long, let us omit the proof of the facts.
Fact 1. Given $X,Y\in H_n$ with $X,Y\ge 0$, $${\rm Im}(X)\supset {\rm Im}(Y)\iff {\rm Ker}(X)\subset {\rm Ker}(Y)\iff X\ge c Y \quad\text{for some} \quad c>0.\tag{4}$$
Fact 2. Given $A\in H_n$ and $B\in H_m$, we have $A\otimes B\in H_{mn}$ and
$$A\ge 0, B\ge 0 \Rightarrow A\otimes B\ge 0.\tag{5}$$
Fact 3. Given $X\in H_{mn}$,
$$X\ge 0 \Rightarrow {\rm tr}_1(X)\ge 0,\ {\rm tr}_2(X)\ge 0.\tag{6}$$
From the definitions $(1)$, $(2)$ and the facts 2 and 3, we have
Fact 4.
Given $Y,Z\in H_n$, if $Y,Z\ge 0$, then for $X=Y+Z$,
$$\mathcal P(X)\ge \mathcal P(Y)+\mathcal P(Z).\tag{7}$$
Fact 5.(special case of Sylvester's law of inertia)
If $X\in H_n$, $X\ge 0$ and $X$ has rank $1\le r\le n$, then there exist $r$ nonzero vectors $v_1,\cdots,v_r\in\mathbb C^n$, such that
$${\rm Im}(X)={\rm span}(v_1,\dots,v_r),\text{ and } X=\sum_{i=1}^rv_iv_i^*.\tag{8}$$
Here $v_i$ is denoted as column vector and $v_i^*$ denotes its conjugate transpose.
Proof of Claim:
Write $X=\sum_{i=1}^rv_iv_i^*$ as in $(8)$. Note that $v_iv_i^*\ge 0$. Then apply $(7)$ inductively we can obtain that
$$\mathcal P(X)\ge\sum_{i=1}^r\mathcal P(v_iv_i^*).$$
Now let us assume that the claim is true when the rank of $X$ is $1$. Under this assumption, from $(4)$ we know that for every $1\le i\le r$, there exists $c_i>0$, such that
$$\mathcal P(v_iv_i^*)\ge c_i v_iv_i^*.$$
It follows that for $c:=\min_{1\le i\le r}c_i>0$,
$$\mathcal P(X)\ge c\sum_{i=1}^r v_iv_i^*=cX\Rightarrow {\rm Im}(\mathcal P(X))\supset {\rm Im}(X).$$
Therefore, it suffices to show that the claim is true when the rank of $X$ is $1$, i.e. $X=vv^*$ for some $v\in\mathbb C^{mn}$.
Denote $v=(w_1,\cdots,w_n)^t$, where $w_1,\cdots,w_n\in\mathbb C^m$ and $t$ means transpose. By definition,
$X=(w_iw_j^*)_{n\times n}$, so by $(1)$,
$${\rm tr_1}(X)=\sum_{i=1}^nw_iw_i^*, \quad{\rm tr_2}(X)=(x_{ij})_{n\times n}, \text{ where } x_{ij}= w_j^*w_i.\tag{9}$$
Note that by $(6)$, for $i=1,2$, ${\rm tr}_i(X)\ge 0$, and hence ${\rm Ker}({\rm tr}_i(X))$ and and ${\rm Im}({\rm tr}_i(X))$ are orthogonal to each other. Then it is easy to see that
$${\rm Ker}(\mathcal P(X))= {\rm Ker}({\rm tr_2}(X))\otimes\mathbb C^m + \mathbb C^n\otimes {\rm Ker}({\rm tr_1}(X)).\tag{10}$$
Therefore, due to $(4)$ and $(10)$, it suffices to show that for $u=u_2\otimes u_1\in \mathbb C^n\otimes\mathbb C^m$, if ${\rm tr}_i(X)u_i=0$ for $i=1$ or $2$, then $Xu=0$.
Denote $u_2=(a_1,\dots,a_n)^t$. Then $u=(a_1u_1,\dots, a_nu_1)^t$, and
$$Xu=\big((\sum_{j=1}^n a_jw_j^*u_1)w_1,\cdots,(\sum_{j=1}^n a_jw_j^*u_1)w_n\big)^t.\tag{11} $$
From $(9)$ we know that
- ${\rm Ker}({\rm tr}_1(X))$ is the orthogonal complement of ${\rm span}(w_1,\dots,w_n)$, so if ${\rm tr}_1(X)u_1=0$, then $w_j^*u_1=0$ for $j=1,\dots,n$, and hence $(11)$ is $0$.
- ${\rm tr}_2(X)u_2=\big((\sum_{j=1}^n a_jw_j^*)w_1,\cdots,(\sum_{j=1}^n a_jw_j^*)w_n\big)^t$, so if ${\rm tr}_2(X)u_2=0$, $(11)$ is $0$.
The proof of claim is completed. $\quad\square$
Best Answer
Notation: Take $A,B$ to be both $m \times n$ matrices and let $e_1^{(n)},\dots,e_n^{(n)}$ denote the standard basis of $\Bbb R^n$.
You're not quite correct; note that $P(A \otimes B)$ will generally have the wrong number of columns. However, it is possible to write $A \circ B = P(A \otimes B)Q^T$ for an appropriate matrices $P,Q$ that satisfy $PP^T = I_m$ and $QQ^T = I_n$. In the case that $m = n$, we will have $P = Q$.
I don't know of any common name for these matrices. However, we can express $P,Q$ as the following sums: $$ P = \sum_{i=1}^m e_{i}^{(m)}[e_i^{(m)} \otimes e_i^{(m)}]^T, \quad Q = \sum_{i=1}^n e_{i}^{(n)}[e_i^{(n)} \otimes e_i^{(n)}]^T. $$