In the case of numbers this is fine because there is in fact a way of copying them beforehand by an operation Delta : A -> A x A, which sends a number a to (a,a), so if given (a,b) you want both their product and the numbers themselves, you need a map A x A -> A x A x A given by (for example)
(id x M x id) . (id x id x Delta) . (Delta x id)
In the case of differential forms, there is no such (natural) map Delta.
To make this point perfectly clear: Whenever you encounter an expression such as "f(x)", you may freely re-use the expression "x" in a different place. This is a matter independent from the category you choose to work with -- for example, the expression "psi \otimes psi" makes sense in the monoidal category of Hilbert spaces.
In some cases you might be able to assume from the start that you have a certain number of equal morphisms x : I -> A "in reserve", for example in linear logic you have the operator ! ("of course") which gives you an arbitrary number of identical morphisms to play with.
In any case this distinction is quite subtle and I understand, why you might assume that I'm simply misunderstanding things. In particular I should emphasize that almost no programming languages work this way, although with some effort you would be able to recast typical cryptographic / numerical code in this language.
It is also really easy that to see for example in the case of addition that indeed information is destroyed, clearly the map
(a,b) -> (a+b,a-b)
has an inverse if the characteristic isn't 2, the subsequent projection to the first factor destroys information. Categories in which that is possible have maps d_A : A -> I.
What tel is pointing out above, is that Delta and d together form a comonoid.
While psi \otimes psi certainly makes sense, the map psi \to psi \otimes psi is not linear and therefore not a morphism (Physicists call this the no cloning theorem for some reason).
The non-existence of a morphism "psi -> psi \otimes psi" and the notion of "destroyed information" that you are discussing in the rest of your post is independent from all of this. If you wish I can elaborate on the "no cloning theorem".