I am thinking about the worst-case space complexity of an algorithm.
Obviously, if $f \in O(nm)$ then $f \in O(n+nm)$. But is the converse true?
$O(m)+O(nm) = O(m+nm) = O(m(1+n)) = O(m)O(1+n) = O(m)O(n) = O(nm)$
If that were true then $O(n+nm) = O(m+nm)$
As mentioned in the question, obviously $O(nm) \subseteq O(n + nm)$ as $n,m \rightarrow \infty $.
The reverse is also true when $n, m$ grow ever larger:
For all $n,m \geq 1 : m +nm \leq nm+nm = 2nm$ and so: $O(m+nm) \subseteq O(nm+nm) = O(2nm) = O(nm)$ as $n,m \rightarrow \infty$.
$O(2nm) = O(nm)$ as $n,m \rightarrow \infty$ is a property of big-O that follows directly from the definition.
With respect to the penultimate line in the question (the long sequence of equalities), there is some unusual notation. To make sense there, the product of two sets must be an element-wise product such as $O(n) \cdot O(m) =\{fg ∣ f∈O(n), g∈O(m)\}$. The ‘$+$’ sign before the first equal sign probably means the Minkowski sum A+B={a+b ∣ a∈A and b∈B} instead of A∪B, but in both cases the first equality would hold.
