I can give an answer for divergence-type operators. Let
$$
Q(u,v)=\int_\Omega a\nabla u\cdot\nabla v\,dx
$$
for $a=(a_{jk})$ with $a_{jk}\in L^\infty_{\mathrm{loc}}(\Omega)$ and $\sum_{j,k}a_{jk}(x)\xi_j\xi_k\geq c|\xi|^2$ for some $c>0$ and all $x\in\Omega$, $\xi\in\mathbb{R}^n$ (indeed, it suffices to assume the strong ellipticity condition only locally). Then $Q$ is a closed form on $D(Q)=\{u\in L^2(\Omega)\mid \int a\nabla u\cdot\nabla u<\infty\}$. If $a_{jk}\in L^\infty(\Omega)$, then $D(Q)=H^1(\Omega)$.

Furthermore, the generator of $Q$ is the divergence-type operator
$$
Lu=\mathrm{div}(a\nabla u)
$$
with Neumann boundary conditions whenever you can make sense of that expression (say, if $a_{jk}\in C^\infty$).

If one restricts $Q$ to $D(Q_0)=\overline{C_c^\infty(\Omega)}^{\lVert\cdot\rVert_Q}$, then the generator is the corresponding divergence-type operator with Dirichlet boundary conditions.

The important property of $Q$ is that it is a Dirichlet form (see [MR92], Sectiion II.2), that is, if $u\in D(Q)$, then $(u\vee 0)\wedge 1\in D(Q)$ and $Q((u\vee 0)\wedge 1)\leq Q(u)$. The same is true for $Q_0$. Essentially, one only needs a sufficiently strong chain rule to see that
$$
Q((u\vee 0)\wedge 1)=\int_\Omega 1_{0\leq u\leq 1} a\nabla u\cdot \nabla u\,dx\leq Q(u).
$$
By abstract theory, this property of $Q$ is equivalent to $0\leq e^{tL}u\leq 1$ for all $0\leq u\leq 1$ (such semigroups are called *sub-Markovian*). This equivalence was probably first stated in [BD58], but without proof. A proof can be deduced from the more general main theorem in [Ouh96] (of course, there are also direct proofs, I just don't have a reference at hand).

Since $(e^{tL})$ is sub-Markovian, it extends to a strongly continuous contraction semigroup on $L^\infty(\Omega)$, and by taking adjoints one also gets a strongly continuous contraction semigroup on $L^1(\Omega)$. Since $^{tL}$ is symmetric on $L^2$, this construction yields the original semigroup on $L^1\cap L^2$.

So, to answer your question:

Yes, the inequality holds, you can choose $C=1$, you do not even have
to restrict to small times, $e^{tL}f$ converges to $f$ in $L^1$-norm as $t\to 0$, and it does not matter whether $\Omega$ is
bounded or not.

**References:**

[MR92] Ma, Röckner. Introduction to the theory of (non-symmetric) Dirichlet forms.

[BD58] Beurling, Deny. Espaces de Dirichlet. I. Le cas élémentaire.

[Ouh96] Ouhabaz. Invariance of closed convex sets and domination criteria for semigroups.