Have you tried the classical Tychonoff's example, $$u(t,x) = \sum_{k = 0}^\infty \frac{g^{(k)}(t) x^{2k}}{(2k)!},$$ with $g(t) = e^{-1/t^\alpha}$ and $\alpha > 1$?

As discussed, for example, in these lecture notes, in this case $$|g^{(k)}(t)| \leqslant \frac{k!}{(\theta t)^k} e^{-1/(2 t^\alpha)} \tag{$\star$}$$ for some $\theta > 0$ (*edit: see below for a proof*), and hence
$$ |u(t, x)| \leqslant \sum_{k = 0}^\infty \frac{k! x^{2k}}{(\theta t)^k(2k)!} e^{-1 / (2 t^\alpha)} \leqslant \sum_{k = 0}^\infty \frac{x^{2k}}{(2 \theta t)^k k!} e^{-1 / (2 t^\alpha)} = e^{x^2/(2 \theta t)-1/(2t^\alpha)} \leqslant e^{x^2 / (2 \theta t)} . $$
If I am not mistaken, this (counter-)example implies a negative question to your answer.

*Edit:* For completeness, here is the proof of ($\star$). This is, of course, completely standard; I reworked it only to convince myself that there is no error in my answer.

The function $g(t) = e^{-1/t^\alpha}$ is holomorphic in the right complex half-plane. By Cauchy's formula (applied to the circle $\Gamma$ centred at $t$ with radius $\theta t$), we have
$$ g^{(k)}(t) = \frac{k!}{2\pi i} \int_\Gamma \frac{g(z)}{(z - t)^{k + 1}} dz = \frac{k!}{2\pi} \int_0^{2\pi} \frac{g(t + \theta t e^{i s})}{(\theta t e^{i s})^k} ds . $$
Therefore,
$$ |g^{(k)}(t)| \leqslant \frac{k!}{2\pi (\theta t)^k} \int_0^{2\pi} |g(t + \theta t e^{i s})| ds . $$
Choose $\theta > 0$ small enough, so that the image of the disk $D(1, \theta)$ under $z \mapsto z^{-\alpha}$ is contained in the half-plane $\Re z > \tfrac{1}{2}$. Since $|g(z)| = e^{-\Re z^{-\alpha}}$, we have $$|g(t + \theta t e^{i s})| = e^{-t^{-\alpha} \Re (1 + \theta e^{i s})^{-\alpha}} \leqslant e^{-t^{-\alpha} / 2} .$$
It follows that $|g^{(k)}(t)| \leqslant \frac{k!}{(\theta t)^k} e^{-1 / (2 t^\alpha)}$, as desired.