Given a bounded linear operator $T$ on a tensor product $H_1 \otimes H_2$, the operator $T$ is not necessarily of the form $T_1 \otimes T_2$, where $T_i:H_i \to H_i$. Even in finite dimensions, $T$ is not necessarily of this form.
Here is a simple counterexample. Let $H_1 = H_2 = \mathbb C^n$. Let $T$ be the "swap" map, the linear extension of
$$
T(x \otimes y)= y \otimes x.
$$
After a little bit of thought, it should become intuitively clear that $T \neq T_1 \otimes T_2$ for any maps $T_1, T_2$; the maps $T_i$ depend only on the space $H_i$, but the map $T$ needs to "work with" both inputs $x$ and $y$ simultaneously.
You can easily obtain a proof by fixing orthonormal bases $(e_i)$ for the Hilbert spaces $H_i$. If $T_1$ corresponds to the matrix $(t_{ij}^1)$ and $T_2$ to the matrix $(t_{kl}^2)$, then $T_1 \otimes T_2$ corresponds to the matrix $(t_{ij}^1t_{kl}^2)$. We will then have,
$$(T_1 \otimes T_2)(e_s \otimes e_t)
=\sum_{is}t_{ij}^1t_{kt}^2 \ e_i \otimes e_k.
$$
If this is to equal $e_t \otimes e_s$, we must have $t_{is}^1 = \delta_{it}$ and $t_{kt}^2=\delta_{ik}$. But these constraints need to hold for every choice of $s$ and $t$ simultaneously and are thus contradictory. (For example, the conditions imposed on $t_{ij}^1$ by insisting that $e_1 \otimes e_2 \mapsto e_2 \otimes e_1$ and also that $e_1 \otimes e_3 \mapsto e_3 \otimes e_1$ are contradictory.)
Another way of thinking about this is as follows. Let $L(H)$ denote the bounded linear operators on a Hilbert space $H$. We have an inclusion $L(H_1) \otimes L(H_2)$ into $L(H_1 \otimes H_2)$ (as described in the first paragraph of your question). If the spaces $H_i$ are finite-dimensional, we even have
$$
L(H_1) \otimes L(H_2) = L(H_1 \otimes H_2).
$$
But not every element in $L(H_1) \otimes L(H_2)$ is an elementary tensor, i.e., of the form $T_1 \otimes T_2$. Even when $H_i$ is finite-dimensional, the most we can ask for in general is that $T$ is of the form
$$
T=\sum_i T_{1,i} \otimes T_{2,i}.
$$
Firstly, it is immediate that $$\operatorname{Im}(T \otimes S) = \operatorname{Im}(T) \otimes \operatorname{Im}(S) \subseteq\overline{\operatorname{Im}(M)}\otimes \overline{\operatorname{Im}(N)} \subseteq \overline{\operatorname{Im}(M\otimes N)}$$
and so $\overline{\operatorname{Im}(T \otimes S)} \subseteq \overline{\operatorname{Im}(M\otimes N)}$
Now, $\operatorname{Im}(T \otimes S) = T \hat \otimes S (E \otimes F)$. But $T \hat \otimes S$ is a continuous map on $E \hat \otimes F$ and so $$T \hat \otimes S(E \hat \otimes F) = T \hat \otimes S( \overline{E \otimes F}) \subseteq \overline{T \hat \otimes S(E \otimes F)}.$$
Therefore $\operatorname{Im}(T \hat \otimes S) \subseteq \overline{\operatorname{Im}(T\otimes S)} \subseteq \overline{\operatorname{Im}(M\otimes N)}$
Best Answer
In the finite-dimensional setting, every operator is Fredholm with index zero. So let's assume $\mathcal H$ is infinite-dimensional.
Firstly, note that if $x\in\ker(T)$ is non-trivial, then $x\otimes y\in\ker(T\otimes S)$ for any $y\in\mathcal H$. Thus $\ker(T\otimes S)$ is infinite-dimensional as long as either $\ker(T)$ or $\ker(S)$ is non-trivial, and therefore $T\otimes S$ is not Fredholm. Similarly, we see that if $\ker(T^*)$ or $\ker(S^*)$ is non-trivial, then $T\otimes S$ is not Fredholm. Thus the only time that $T\otimes S$ is Fredholm is when $T$ and $S$ are both invertible, in which case $T\otimes S$ is invertible, and the index of all three operators is zero.