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
$$
T=\sum_i T_{1,i} \otimes T_{2,i}.
$$
The approach based on decomposing $S_1$ and $S_2$ as linear combinations of unitary operators does prove that $S_1\otimes S_2$ is
bounded but, as far as I can remember, does not directly give the sharp estimate
$$
\|S_1\otimes S_2\|\leq \|S_1\|\|S_2\|,
$$
since the coefficients in the aforementioned linear combinations have little relationship with the norm of the
operators involved.
Here is a different approach, completely avoiding unitaries, and which gives the desired norm estimate.
Lemma 1. If $T:H_1\to H_2$ is a bounded linear operator between Hilbert spaces $H_1$ and $H_2$, and given $x_1,x_2,\ldots ,x_n\in H_1$, let $A$ and $B$ be the $n\times n$
matrices defined by
$$
a_{ij} = \langle T(x_i),T(x_j)\rangle ,
\quad \text{and} \quad
b_{ij} = \langle x_i,x_j\rangle .
$$
Then $A\leq \|T\|^2 B$.
Proof. Given scalars $\lambda _1,\lambda _2,\ldots ,\lambda _n$, we have
$$
\sum_{i,j}\lambda _i\bar\lambda _ja_{ij} =
\sum_{i,j}\lambda _i\bar\lambda _j\langle T(x_i),T(x_j)\rangle =
\sum_{i,j}\langle T(\lambda _ix_i),T(\lambda _jx_j)\rangle = $$$$ =
\left\langle \sum_iT(\lambda _ix_i),\sum_jT(\lambda _jx_j)\right\rangle =
\left\|T\left(\sum_i \lambda _ix_i\right)\right\|^2 \leq $$$$ \leq
\|T\|^2\left\|\sum_i \lambda _ix_i\right\|^2 =
\|T\|^2 \sum_{i,j}\langle \lambda _ix_i,\lambda _jx_j\rangle = $$$$ =
\|T\|^2 \sum_{i,j}\lambda _i\bar\lambda _j\langle x_i,x_j\rangle =
\|T\|^2 \sum_{i,j}\lambda _i\bar\lambda _jb_{ij}.
$$
QED
Lemma 2. If $T:H_1\to H_2$ is a bounded linear operator between Hilbert spaces $H_1$ and $H_2$, and $K$ is another Hilbert
space, then there is a unique linear operator
$$
T\otimes I:H_1\otimes K\to H_2\otimes K,
$$
such that
$$
T(x\otimes y) = T(x)\otimes y, \quad \forall x\in H_1, \quad \forall y\in K.
$$
Moreover $\|T\otimes I\|\leq \|T\|$.
Proof. By replacing $T$ with $T/\|T\|$, we may assume that $\|T\|=1$.
The crucial step in the proof is then showing the inequality
$$
\left\| \sum_i T(x_i)\otimes y_i \right\| \leq \left\| \sum_i x_i\otimes y_i \right\|,
$$
whenever $x_1,x_2,\ldots ,x_n\in H_1$, and $y_1,y_2,\ldots ,y_n\in K$.
In this case,
observe that the matrices $A$ and $B$ defined in Lemma (1) satisfy $A\leq B$, whence $B-A\geq 0$,
so we may find an $n\times n$ scalar matrix $C$ such that
$B-A=CC^*$. Therefore, for every $i$ and $j$ one has that
$$
\langle x_i, x_j\rangle = \langle T(x_i), T(x_j)\rangle + \sum_k c_{ik}\bar c_{j k}.
$$
Multiplying by $\langle y_i, y_j\rangle $ and summing on $i$ and $j$, we get
$$
\sum_{i, j}\langle x_i, x_j\rangle \langle y_i, y_j\rangle = \sum_{i, j}\langle T(x_i), T(x_j)\rangle \langle y_i, y_j\rangle + \sum_{i, j}\sum_k c_{ik}\bar c_{jk}\langle y_i, y_j\rangle ,
$$
which means that
$$
\left\|\sum_i x_i\otimes y_i \right\|^2 = \left\|\sum_i T(x_i)\otimes y_i \right\|^2 + \sum_{i, j}\sum_k c_{ik}\bar c_{j k}\langle y_i, y_j\rangle ,
$$
so it suffices to show that the last sum above is nonnegative. But this is easy because
$$
\sum_{i, j}\sum_k c_{ik}\bar c_{j k}\langle y_i, y_j\rangle =
\sum_k \left\langle \sum_i c_{ik}y_i,\sum_jc_{j k}y_j\right\rangle =
\sum_k \left\|\sum_i c_{ik}y_i\right\|^2 \geq 0.
$$
QED
Adopting the notation used by the OP, we may now easily prove that $S_1\otimes S_2$ is bounded, and indeed that
$\|S_1\otimes S_2\|\leq \|S_1\|\|S_2\|$, as follows
$$
\|S_1\otimes S_2\| = \|(S_1\otimes I)(I\otimes S_2)\| \leq \|S_1\otimes I\|\|I\otimes S_2\| \leq \|S_1\|\|S_2\|.
$$
Best Answer
$\newcommand{\norm}[1]{\left\|{#1}\right\|} \newcommand{\ip}[1]{\left\langle{#1}\right\rangle} \newcommand{\abs}[1]{\left|{#1}\right|}$Let $S \in B(H_1)$, $T \in B(H_2)$; the claim is that $\norm{S \otimes T} \leq \norm{S}\norm{T}$.
Let me first show that $S \otimes I$ is bounded with $\norm{S \otimes I} \leq \norm{S}$; the same proof, mutatis mutandis, will show that $I \otimes T$ is bounded with $\norm{I \otimes T} \leq \norm{T}$. Since the algebraic tensor product $H_1 \odot H_2$ is dense in $H_1 \otimes H_2$, it suffices to show that $\norm{(S \otimes I)v} \leq \norm{S}\norm{v}$ for any $v \in H_1 \odot H_2$.
So, let $v = \sum_{k=1}^N x_k \otimes y_k \in H_1 \odot H_2$; by performing Gram--Schmidt orthogonalisation on $\{y_k\}$ and expressing the $y_k$ in terms of the resulting orthonormal basis for $\operatorname{span}\{y_k\}$, we may assume without loss of generality that $\{y_k\}$ is orthonormal. On the one hand, it follows that $\{x_k \otimes y_k\}$ is orthogonal, so that $$ \norm{v}^2 = \norm{\sum_{k=1}^N x_k \otimes y_k}^2 = \sum_{k=1}^N \norm{x_k \otimes y_k}^2 = \sum_{k=1}^N \norm{x_k}^2. $$ On the other hand, since $(S \otimes I)(x_k \otimes y_k) = Sx_k \otimes y_k$, it follows that $\{Sx_k \otimes y_k\}$ is also orthogonal, so that by the same computation, mutatis mutandis, $$ \norm{(S \otimes I)v}^2 = \sum_{k=1}^N \norm{S x_k}^2 \leq \sum_{k=1}^N \norm{S}^2 \norm{x_k}^2 = \norm{S}^2 \sum_{k=1}^N \norm{x_k}^2 = \norm{S}^2\norm{v}^2. $$ Thus, $\norm{(S \otimes I)v} \leq \norm{S}\norm{v}$, as required.
Now, observe that since $(S \otimes T) = (S \otimes I)(I \otimes T)$ on $H_1 \odot H_2$, it follows by the boundedness of $S \otimes I$ and $I \otimes T$ that $S \otimes T$ is also bounded with norm $$ \norm{S \otimes T} \leq \norm{S \otimes I}\norm{I \otimes T} \leq \norm{S}\norm{T}, $$ as required.