The answer is no in general - but this is a fairly subtle issue. First, let's go over why this is "almost true".
Given $e:X \to N$ I denote by $X_0$ and $X_{>0}$ the pullback of $\{0\}$ and $N_{>0}$ along $e$.
So, given $N$ a parametrized NNO, and $f,g :X \to Y$ two functions, what you can easily do using the NNO property, is construct a function $h:X \times N \to Y$ such that $h(x,0)= f(x)$ and $h(x,Sn)=g(x)$.
And if on top of this you have a map $e:X \to N$ then you can define $h'(x)=h(x,e(x))$ which morally is defined $h'(x) = f(x)$ if $e(x)=0$ and $h'(x)=g(x)$ if $e(x)>0$. More precisely, its restriction to $X_0$ and $X_{>0}$ will coincide respectively with the restriction of $f$ and $g$.
But that doesn't quite give you a pushout because the maps $f$ and $g$ are defined on the whole of $X$ and not on $X_0$ and $X_{>0}$. If you have a way to somehow extend a map only defined on $X_0$ and $X_{>0}$ to a map defined on the whole of $X$, then this construction would show the result (at least the existence part). For example, I claim that this method shows:
Proposition: Let $e:X \to N$. Assume there are two maps $u_0:X \to X_0$ and $u_{>0} :X \to X_{>0}$. Then $X = X_0 \coprod X_{>0}$
For example, if both $X_0$ and $X_{>0}$ have global sections (map from $1$) then the proposition can be applied with $u_0$ and $u_{>0}$ two constant maps.
The general idea of the proof is that using the discussion above you can modify $u_0$ and $u_{>0}$ into maps $\tilde{u_0}$ and $\tilde{u_{>0}}$ so that they restrict to the identity on $X_0$ and this provide you with a canonical way of extending functions on $X_0$ and $X_{>0}$ to the whole of $X$ to apply the discussion above. This only proves the existence part of the coproduct property, not the uniqueness - but there is a neat trick to deduce uniqueness:
Lemma : Let $i:A \to C$ and $j:B \to C$ and assume that the induced natural transformation $Hom(C,Y) \to Hom(A,Y) \times Hom(B,Y)$ admit a section, which is natural in $Y$ and sends the pair $(i,j)$ to the identity of $C$. Then $C$ is the coproduct of $A$ and $B$ (with the maps $i$ and $j$).
Indeed, under the assumption of the lemma, the functor $Hom(A,Y) \times Hom(B,Y)$ is a retract of $C$, by a certain map $P: Hom(C,Y) \to Hom(C,Y)$, which by the Yoneda lemma corresponds to a map $P:C \to C$, but the second assumption give you that $P$ is the identity.
To deduce the proposition, just observe that the way we build a map $X \to Y$ out of two maps $X_0 \to X $ and $X_{>0} \to X$ is natural in $Y$, and gives the identity of $X$ if one start form the two inclusion map (for this we need to use the precise way the function $\tilde{u_0}$ and $\tilde{u_{>0}}$ were defined using the NNO).
Now, a counterexample.
So the general idea is we want to start from a situation where one of the two fiber has no map from 1 - that's not quite sufficient but this is necessary. I'm starting from the following assumption. Let $C$ be a category such that:
$C$ has all finite limits and is extensive (all coproducts are universal and disjoint). I don't really need extensivity, I think I only need a (maybe strict) initial object - but most categories with NNO are extensive anyway.
$C$ has a parametrized NNO.
There exists a function $f: N \to N$ such that $f$ is not the constant equal to zero function but for all map $p: 1 \to N$ the composite $fp$ is equal to $0$.
I would recommend just assuming these exist. But a typical example of this situation (and to be honest the only ones I know) is to take $C$ to be the free topos (with NNO) or the free (extensive) cartesian closed category with NNO, or the free extensive category with NNO. In each case you can take $f$ to be a primitive recursive function which sends an integer $n$ to $1$ if $n$ is a code of a proof in the theory of elementary topos with NNO there exists a map $ * \to \emptyset$ - of course replacing elementary topos with NNO, by extensive (cartesian closed) with NNO in the other two cases.
In these categories, the only maps $1 \to N$ are the standard integer, so as such a proof of course does not exist, all global section are sent to $0$. But the point is that because of Gödel's theorem (applied to the recursively enumerable theory of say elementary topos with NNO), that function $f$ isn't the function equal to $0$.
Ok, so of course $C$ is extensive, so it isn't our counterexample on the nose.
Take $P$ to be the full subcategory of $C$ of objects $X$ that are either the initial object or admit at least one function $1 \to X$.
I claim that $P$ is a coreflective subcategory of $C$. Indeed, for any object $Y$ of $C$, then either it has no map $1 \to Y$ - in which case there is only one map $X \to Y$ with $X \in P$ given by the unique map from the initial object - Or there are some maps $1 \to Y$ in which case $Y \in P$. So in both case, there is a universal map $X \to Y$ with $X \in P$.
In particular, the category $P$ has all finite limits (by computing them in $C$ and applying the reflection). It is closed under product and coproduct in $C$ and contains the NNO of $C$, so it has a parametrized NNO.
Now consider our function $f:N \to N$, we assumed above. The fiber of $f$ over $N_{>0}$ has no map from $1$, so in $P$ it becomes the initial object. The fiber of $f$ over $0$ cannot be isomorphic (by the canonical mao) to $N$ because otherwise, we would have $f=0$. So the coproduct of the two is simply the fiver over $0$ - which is not isomorphic to $N$ (well - in fact, it will likelt be isomorphic to $N$ - but I mean not isomorphic by the canonical map).
