Let's do this in $n$ variables, and over $\mathbb C$. The polynomials $P$ such that $\ker \partial P_0 \subset \ker \partial P$ form an ideal $J$ in ${\mathbb C}[z_1,\ldots,z_n]$ which contains $P_0$. The question is whether this ideal is generated by $P_0$.
For $f(z_1,\ldots,z_n) = \exp(\sum_j a_j z_j)$, where $a_j \in \mathbb C$, we have $(\partial P_0) f = P_0(a_1,\ldots,a_n) f$, so $f \in \ker \partial P_0$ iff $P_0(a_1,\ldots,a_n) = 0$. Let $V_0 = \{(a_1,\ldots,a_n) \in {\mathbb C}^n: P_0(a_1,\ldots,a_n)=0\}$, and let $J(V_0)$ be the ideal of ${\mathbb C}[z_1,\ldots,z_n]$ consisting of polynomials $P$ that are $0$ on $V_0$. This contains $J$. By Hilbert's Nullstellensatz, $J_0$ is actually the radical of the ideal generated by $P_0$. This will be the ideal generated by $P_0$ if $P_0$ is the product of distinct irreducible polynomials. And then $J$ is generated by $P_0$.
EDIT: If $P_0$ has a factor that is the $k$'th power of a polynomial, the situation is less clear to me. I think you have to also consider functions $f$ that are products of a polynomial and an exponential.
EDIT: In fact the statement is true for all polynomials $P_0$.
Suppose $\ker \partial P_0 \subset \ker \partial P$. Let $P_0 = Q^m R$ where $Q$ is irreducible and $\gcd(Q,R) = 1$. I want to show that $P$ is divisible by $Q^m$. In fact let $P = Q^k S$ where $\gcd(Q,S) = 1$, so I need to show $k \ge m$. Take $a = (a_1, \ldots, a_n) \in {\mathbb C}^n$ such that $Q(a_1,\ldots,a_n) = 0$ while $R(a_1,\ldots,a_n) \ne 0$ and $S(a_1,\ldots,a_n) \ne 0$. Note that for any polynomial $p$, $\partial p (g \ e^{a \cdot z}) = e^{a \cdot z} \partial(\tau_a p)(g)$ where $a \cdot z = a_1 z_1 + \ldots + a_n z_n$ and $(\tau_a p)(x_1,\ldots,x_n) = p(x_1 + a_1, \ldots, x_n + a_n)$. $\tau_a$ is an automorphism of the ring ${\mathbb C}[x_1,\ldots,x_n]$, and $\ker \partial P_0 \subset \ker \partial P$ iff $\ker \partial (\tau_a P_0) \subset \ker \partial (\tau_a P)$. Thus without loss of generality we can assume $a = (0,\ldots, 0)$.
Now consider the linear subspace $X_d$ of polynomials in ${\mathbb C}[x_1,\ldots,x_n]$ of degree $\le d$ (for the rest of this paragraph, all operators are considered as acting on this space, which is invariant under all constant-coefficient differential operators). Take $d$ large enough that $\partial Q^m$ is not $0$. Since $Q(0,\ldots,0) = 0$, if $g$ has degree $d \ge 0$ then $\partial Q(g)$ has degree less than $d$, and $\partial Q^{d+1}(g) = 0$. Thus $\partial Q$ is nilpotent. On the other hand, $\partial S$ is of the form $c I + N$, where $c = S(0,\ldots,0) \ne 0$ and $N$ is nilpotent, and therefore $\ker \partial S = \{0\}$. So we have $\ker \partial Q^m \subset \ker \partial Q^k$. Now take $g$ such that $\partial Q^{m-1}(g) \ne 0$, and let $j$ be the least nonnegative integer such that $\partial Q^{m+j}(g) = 0$ (it exists since $Q$ is nilpotent).
Thus $\partial Q^{j}(g)$ is in $\ker \partial Q^m$ but not in $\ker \partial Q^{m-1}$, so $k \ge m$ as required.