Let the elements we remove be denoted by $s^j=(s_1^j,s_2^j,\ldots,s_k^j)$ for $j=1,2,\ldots,n$. We are interested in the set $$(S_1\times S_2\times\ldots\times S_k)\setminus\{s^1,s^2,\ldots,s^n\}.$$
Claim. A (possibly very crude) upper bound is given by $\frac{n^k-1}{n-1}$.
Proof. We will prove this by induction on $k$.
For $k=1$, there is no problem. The set $S_1\setminus\{s^1,s^2,\ldots,s^n\}$ is a Cartesian product with one factor, so we are done: the upper bound is $1$ in this case.
For $k\to k+1$, note that we can rewrite the set as follows: $$\begin{align}&(S_1\times S_2\times\ldots\times S_{k+1})\setminus\{s^1,s^2,\ldots,s^n\}=\\&=(S_1\times S_2\times\ldots\times S_{k+1})\setminus\{s^1,s^2,\ldots,s^n\}\\&=\bigcup_{x\in S_{k+1}}(S_1\times S_2\times\ldots\times S_k\times\{x\})\setminus\{s^1,s^2,\ldots,s^n\}\\&=\left(\bigcup_{j=1}^n(S_1\times S_2\times\ldots\times S_k\times\{s_{k+1}^j\})\setminus\{s^1,s^2,\ldots,s^n\}\right)\cup(S_1\times S_2\times\ldots\times (S_{k+1}\setminus\{s_{k+1}^1,\ldots,s_{k+1}^n\})),\end{align}$$
which is (if we omit the sets that appear more than once in the first union) a disjoint union of at most $n$ sets of the form $$(S_1\times S_2\times\ldots\times S_k\times\{s_{k+1}^j\})\setminus\{s^1,s^2,\ldots,s^n\},$$ and a single Cartesian product of $k+1$ sets. By the induction hypothesis, each set $$(S_1\times S_2\times\ldots\times S_k\times\{s_{k+1}^j\})\setminus\{s^1,s^2,\ldots,s^n\}$$ can be written as a disjoint union of at most $\frac{n^k-1}{n-1}$ Cartesian products. Therefore our set can be written as a disjoint union of at most $n\frac{n^k-1}{n-1}+1 = \frac{n^{k+1}-1}{n-1}$ Cartesian products (with $k+1$ factors each, of course). $\square$
Note that finding better bounds might require more sophisticated combinatorial methods.
Best Answer
There indeed do exist many well-known laws for Cartesian products, such as the following:
$(A \cap B) \times (C \cap D) = (A \times C) \cap (B \times D),$
$A \times (B \cap C) = (A \times B) \cap (A \times C)$ (distributivity of intersection),
$A \times (B \cup C) = (A \times B) \cup (A \times C)$ (distributivity of union),
$A \times (B \backslash C) = (A \times B) \backslash (A \times C)$ (distributivity of set difference),
and various other laws. For more, see here: https://en.wikipedia.org/wiki/Cartesian_product#Most_common_implementation_(set_theory)