As suggested in the question, $P_t$ need *not* be a well defined operator on $L^\infty(W,\mu)$. That is, $F$ can be zero $\mu$-almost everywhere, but $P_tF$ is nonzero on a set of positive $\mu$-measure.

For example, take the Banach space $W$ to be $\ell^2$. Let $X_0,X_1,\cdots$ be an IID sequence of normal random variables with zero mean and unit variance, defined on some probability space $(\Omega,\mathcal{F},\mathbb{P})$. Then, set $X=(X_0,2^{-1}X_1,\ldots,2^{-n}X_n,\ldots)\in\ell^2$ (a.s.), and let $\mu$ be the measure of $X$.
Define the measurable function $G\colon\ell^2\to\mathbb{R}$ by
$$
G(c)=\begin{cases}
\lim_{n\to\infty}\frac1n\sum_{k=0}^{n-1}4^kc_k^2,&\textrm{if the limit exists},\cr
0,&\textrm{otherwise}.
\end{cases}
$$
By the strong law of large numbers,
$$
G(tX)=t^2\lim_{n\to\infty}\frac1n\sum_{k=0}^{n-1}X_k^2=t^2\textrm{ (a.s.)}
$$
You can now define $F\colon\ell^2\to\mathbb{R}$ by $F(c)=0$ when $G(c)=1$ and $F(c)=1$ otherwise. Then, $F(X)=0$ (a.s.) and $F(tX)=1$ for $t > 1$ (a.s.), so $F=0$ $\mu$-almost everywhere. However, the measure $\int P_t(x,\cdot)d\mu(x)$ is the same as the distribution of $(1+t)^{1/2}X$. Therefore, $P_tF=1$ $\mu$-almost everywhere for each $t > 0$.