= Solution
If $u\in\mathcal E'(\mathbb R)$ solves $P(D)u=v$, Fourier transformation gives
$$
P(z)\widehat u(z)=\widehat v(z).
$$
The <Paley–Wiener–Schwartz theorem> makes $\widehat u$ entire, so $\widehat v/P$ is entire.
Conversely, suppose $F(z)=\widehat v(z)/P(z)$ is entire. Polynomial division estimates away from the finitely many zeros of $P$, together with the maximum principle on fixed disks around those zeros, show that $F$ retains a Paley--Wiener--Schwartz bound, with only the polynomial exponent changed. The converse theorem therefore gives $u\in\mathcal E'(\mathbb R)$ with $\widehat u=F$. Then $P(D)u=v$. Thus
$$
\boxed{\exists u\in\mathcal E':P(D)u=v
\iff \widehat v/P\text{ is entire}}.
$$
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