Why doesn't Kallenberg include the condition $T_{0} = I$ in his definition of a Feller semigroup?

Question

I'm a bit confused about Kallenberg's definition of a Feller semigroup in Foundations of Modern Probability, 3rd ed. The setup: $S$ is a locally compact, separable metric space and $C_{0} = C_{0}(S)$ is the (Banach) space of continuous functions on $S$ that vanish at infinity, with the supremum norm.

The definition: a semigroup of positive contraction operators $T_{t}$ on $C_{0}$ is a Feller semigroup if

1. $T_{t}C_{0} \subseteq C_{0}$
2. $T_{t}f(x) \to f(x)$ as $t \to 0$ for all $f \in C_{0}$, $x \in S.$

My question is: why is the condition $T_{0} = I$ (the identity operator) not included in the definition?

I have seen other sources define a Feller semigroup with the condition $T_{0} = I$ included (for instance, Klenke's Probability Theory and Eberle's Markov processes) and I can't tell what difference it makes.

One possibility is that Kallenberg implicitly includes $T_{0} = I$ in his definition of a semigroup along with the usual condition $T_{s}T_{t} = T_{s+t}$, which some authors in functional analysis seem to do. But this isn't laid out anywhere in the text (that I can find.)

The reason this is troubling me is that I'm unsure what the lack of "$T_{0} = I$" means for the transition kernels that correspond to $T_{t}$. If $S$ is compact and $T_{t}$ is a conservative ($T_{t}1 = 1$) Feller semigroup on $C_{0}$, then there exists a unique semigroup of transition kernels $\mu_{t}$ on $S$ satisfying

$$T_{t}f(x) = \int f(y)\,\mu_{t}(x,dy)\qquad \forall x \in S, f \in C_{0} \qquad (1)$$

If we interpret $\mu_{t}(x,B)$ as "the probability of transitioning from $x$ to the measurable set $B$ in time $t$," then it seems like the only reasonable possibility is for

$$\mu_{0}(x,B) = \begin{cases} 1 \quad & x \in B \\ 0 \quad & \text{otherwise} \end{cases}$$

i.e., $\mu_{0}(x,\cdot)$ is the point measure on $x$. In this case, from (1) we have $T_{0}f(x) = f(x)$ for all $x, f$, i.e., $T_{0} = I$.

But if $T_{0}$ needn't be the identity in general, how can we then make sense of $\mu_{0}$? Wouldn't it have to assign some nonzero probability to making a transition in $0$ time?