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.

**[Updated 2022-12-06]**
Second, consider the category $\mathcal{D}$ whose objects are the finite powers of $\mathbb{N}$, like before, and whose morphisms $\mathbb{N}^m \to \mathbb{N}^n$ are maps of the form
$$(x_1, \ldots, x_m) \mapsto (x_{r(1)} + k_1, \ldots, x_{r(n)} + k_n)$$
where $r : \{1, \ldots, n\} \to \{1, \ldots, m\}$ is any map and $k_1, \ldots, k_n \in \mathbb{N}$. In words, the morphisms in $\mathcal{D}$ are projections composed with addition of a constant. One verifies easily that these form a category.

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.