Let $H$ be a Hilbert space and suppose that the densely defined linear operator $L:D(L) \subseteq H \to H$ is the generator of a $C_0$ semigroup $e^{Lt}:H \to H$. Suppose further that $S \subseteq D(L)$ is a dense topological vector space such that $L$ maps $S$ continuously into itself. Is it true that if one takes $h \in S$, then $e^{Lt}h \in S$ for all $t > 0$?
A concrete version of the question would be something like: Does the heat semigroup map the Schwarz space into itself?
The heat semigroup does map the Schwartz space into itself, because the heat kernel $(2\pi t)^{-n/2} e^{-|x|^2/2t}$ is a Schwartz function, and the Schwartz space is closed under convolution.
However, your statement is false in general: take again $H=L^2(\mathbb{R}^n)$, $L$ the Laplacian with its domain $H^2(\mathbb{R}^n)$, and $S = C^\infty_c(\mathbb{R}^n)$ with its usual topology. Then $S$ is invariant under $L$, and the restriction of $L$ to $S$ is continuous (in the topology of $S$). But $S$ is clearly not invariant under $e^{tL}$, because solutions of the heat equation are not compactly supported.