It seems that most small simple-graphs cannot be made by multiplying smaller simple graphs using cartesian multiplication, with the exception of products involving the trivial single node graph.
A more concise way of saying this could be that most have no cartesian quotient that is also a simple-graph. Or further more that most simple-graphs are (cartesian) prime.
However a large proportion of single digit natural numbers are prime while the density of prime numbers tends towards 0 as the magnitude of the numbers being considered tends towards infinity.
Does the density of "cartesian prime" simple-graphs also decrease as the number of nodes in the graph increases?
There's $$2^{\binom{n}{2}}$$ labeled $n$-vertex graphs, which has exponent $\sim n^2/2$, and $$\prod_{\substack{d \text{ divides } n\\d \not\in \{1,n\}}} 2^{\binom{d}{2}+\binom{n/d}{2}} \leq n2^{2\binom{n/2}{2}} \leq 2^{n^2/4+\log_2(n)}$$ labeled $n$-vertex Cartesian product graphs, which has exponent $\sim n^2/4$ (excluding things like $K_1 \Box G$).
Thus, almost all graphs are "Cartesian primes".
We shouldn't expect primes and Cartesian primes to behave the same: the number of numbers in $\{1,2,\ldots,n\}$ grows linearly with $n$, whereas the number of $n$-vertex graphs grows exponential polynomially. Their behaviors are probably largely unrelated.
By the way, for unlabelled graphs, the number of $n$-vertex graphs is no more than $$ \frac{1}{n!}2^{\binom{n}{2}} \sim 2^{n^2/2-n\ln n+n-O(\ln n)} $$ by Stirling's Approximation, which doesn't change much.