I give two examples of categories with finite products and simple NNO. In the first example the simple NNO is also a parameterized NNO, while in the second example it is not. Although it is difficult to understand your question, I believe the examples should clarify matters.
First, consider the category $\mathcal{C}$ whose objects are the finite powers of $\mathbb{N}$, namely $\mathbb{N}^0$, $\mathbb{N}^1$, $\mathbb{N}^2$, ... and morphisms are set-theoretic functions $f : \mathbb{N}^k \to \mathbb{N}^m$. This category clearly has finite products, is not cartesian-closed because there are too many morhisms $\mathbb{N} \to \mathbb{N}$, and it has a parameterized NNO, namely the obvious one.
Second, consider the category $\mathcal{D}$ whose objects are the finite powers of $\mathbb{N}$, like before, and whose morphisms are as follows:
- Morphisms $\mathbb{N}^k \to \mathbb{N}^m$ with $m \neq 1$ are all set-theoretic functions.
- Morphisms $\mathbb{N}^0 \to \mathbb{N}^1$ are all set-theoretic functions, i.e., for each natural number there is one.
- Morphisms $\mathbb{N}^k \to \mathbb{N}^1$ with $k \neq 0$ are all set-theoretic functions $f : \mathbb{N}^k \to \mathbb{N}$ for which there exists a projection $\pi_j : \mathbb{N}^k \to \mathbb{N}$ and $g : \mathbb{N} \to \mathbb{N}$ such that $f = g \circ \pi_j$.
In other words, in $\mathcal{D}$ every function into $\mathbb{N}$ depends on only one of its parameters (exercise: prove that these are closed under composition.) The category $\mathcal{D}$ has finite products and a simple NNO, namely the obvious one, but no parameterized NNO. If it did, we could construct addition ${+} : \mathbb{N}^2 \to \mathbb{N}$ as a morphism in the category.