In H. Cartan – Differential Calculus (1971) p. 29 he investigates differentiating a bi-linear function $f: E_1 \times E_2 \to F$ where $E_1, E_2, F$ are Banach spaces and $E_1 \times E_2$ the product (presumably Cartesian).
He claims $E_1 \times E_2$ to be a Banach space with the obvious rules of addition and scalar multiplication.
I think this might be OK if $E_1, E_2$ are one-dimensional, but not otherwise. To be algebraically complete mustn't he instead use the tensor product $E_1 \otimes E_2$ ?
Since for $E_1, E_2$ at least two dimensional with bases $\{u_1, u_2\}, \{v_1, v_2\}$ there is a very clear counterexample….
$(u_1, v_1), (u_2, v_1), (u_1, v_2), (u_2, v_2) $ are elements of $E_1 \times E_2$.
But then $(u_2, v_1) + (u_1, v_2) + (u_2, v_2) $ is not of the form $(u, v)$ and so not in $E_1 \times E_2$, i.e. $E_1 \times E_2$ is not algebraically closed under addition.
I may have mixed up some concepts in the above.
It seems that $E_1 \times E_2$ with addition and scalar multiplication as noted by @JohnHughes is the direct sum of $E_1, E_2$ and nothing to do with the tensor product.
And then as noted by @JoonasIlmavirta $(u_2, v_1) + (u_1, v_2) + (u_2, v_2) = (u_1 + 2.u_2, v_1 + 2.v_2)$.
Best Answer
The sum you gave is $$ (u_2, v_1) + (u_1, v_2) + (u_2, v_2) = (u_2+u_1+u_2,v_1+v_2+v_2) $$ and this is an element of $E_1\times E_2$ because $u_2+u_1+u_2\in E_1$ and $v_1+v_2+v_2\in E_2$. The vector is indeed of the form $(u,v)$, where $u=u_2+u_1+u_2$ and $v=v_1+v_2+v_2$.
This has nothing do to with Banach spaces. This is all about the concept of a product of two vector spaces as another vector space.