(This is rather late, but I'll answer the question anyways.) No, it is not true that $\mathscr S_h'$ and $\mathscr P$ are (topological) complements in $\mathscr S'$ (in fact, they are not even linear complements). Suppose it were the case that every tempered distribution $f$ admitted a (unique) decomposition of the form $f= g+P$ for some $g \in \mathscr S_h'$ and some $P \in \mathscr P$. Fix a Schwartz function $\varphi$ with $\varphi(0) \neq 0$. Then, for any $f \in \mathscr S'$, $$ \langle f, \dot S_j \varphi \rangle = \langle \dot S_j g , \varphi \rangle + \langle \dot S_j P, \varphi \rangle \to 0 + \langle P, \varphi \rangle $$ as $j \to - \infty$. This implies that the sequence $( \dot S_j \varphi)_{j \leq 0}$ is Cauchy for $\sigma(\mathscr S, \mathscr S')$. Now, $\mathscr S$ is a semi-Montel space (much more is true, but this is enough for our purposes), so $\sigma(\mathscr S, \mathscr S')$ is quasi-complete (hence sequentially complete), and $\sigma(\mathscr S, \mathscr S')$ convergent sequences also converge in the usual Fréchet topology on $\mathscr S'$. As the Fourier transform is a linear homeomorphism on $\mathscr S$ (with its Fréchet topology), we see that the sequence $(\hat \Phi(2^{-j} \cdot) \hat \varphi)_{j \leq 0}$ converges in $\mathscr S$ as $j \to - \infty$. But, $$ \partial_1 ( \hat \Phi(2^{-j} \cdot) \hat \varphi ) = 2^{-j} (\partial_1 \hat \Phi)(2^{-j} \cdot) \hat \varphi + \hat \Phi(2^{-j} \cdot) \partial_1 \hat \varphi.$$ The second term remains bounded in $L^{\infty}$ as $j \to - \infty$. Take some $x_0$ such that $\partial_1 \hat \Phi(x_0) \neq 0$. Then, $$ |2^{-j} \partial_1 \hat \Phi(x_0) \hat \varphi(2^j x_0) | \to \infty$$ as $j \to - \infty$, resulting in a contradiction.

(This is rather late, but I'll answer the question anyways.) No, it is not true that $\mathscr S_h'$ and $\mathscr P$ are (topological) complements in $\mathscr S'$ (in fact, they are not even linear complements). Suppose it were the case that every tempered distribution $f$ admitted a (unique) decomposition of the form $f= g+P$ for some $g \in \mathscr S_h'$ and some $P \in \mathscr P$. Fix a Schwartz function $\varphi$ with $\varphi(0) \neq 0$. Then, for any $f \in \mathscr S'$, $$ \langle f, \dot S_j \varphi \rangle = \langle \dot S_j g , \varphi \rangle + \langle \dot S_j P, \varphi \rangle \to 0 + \langle P, \varphi \rangle $$ as $j \to - \infty$. This implies that the sequence $( \dot S_j \varphi)_{j \leq 0}$ is Cauchy for $\sigma(\mathscr S, \mathscr S')$. Now, $\mathscr S$ is a semi-Montel space (much more is true, but this is enough for our purposes), so $\sigma(\mathscr S, \mathscr S')$ is quasi-complete (hence sequentially complete), and $\sigma(\mathscr S, \mathscr S')$ convergent sequences also converge in the usual Fréchet topology on $\mathscr S$. As the Fourier transform is a linear homeomorphism on $\mathscr S$ (with its Fréchet topology), we see that the sequence $(\hat \Phi(2^{-j} \cdot) \hat \varphi)_{j \leq 0}$ converges in $\mathscr S$ as $j \to - \infty$. But, $$ \partial_1 ( \hat \Phi(2^{-j} \cdot) \hat \varphi ) = 2^{-j} (\partial_1 \hat \Phi)(2^{-j} \cdot) \hat \varphi + \hat \Phi(2^{-j} \cdot) \partial_1 \hat \varphi.$$ The second term remains bounded in $L^{\infty}$ as $j \to - \infty$. Take some $x_0$ such that $\partial_1 \hat \Phi(x_0) \neq 0$. Then, $$ |2^{-j} \partial_1 \hat \Phi(x_0) \hat \varphi(2^j x_0) | \to \infty$$ as $j \to - \infty$, resulting in a contradiction.