Let $X$ be a complete normed space, $T(t):t\ge 0$ a family of bounded operators on $X$, then $T(t)T(s)=T(t+s)$, $T(0)=I$ and $T(t)x \to x$ as $t\to 0+$ implies $t\mapsto T(t)x$ is continuous for each $x\in X$.
The assumption directly says that $T(t)x$ is continuous at $t=0$. Also, for any $t\in [0,\infty)$, $T(t+)x=\lim_{s\downarrow 0} T(t+s)x=\lim_{s\downarrow 0}T(t)T(s)x = T(t)x$ so the right limit exists and equals $T(t)x$.
But I don't see a way to argue left limit. The main problem I find is that $T(t)$ is not defined for $t<0$, so we can't really do same thing as above. If $s<t$ an $s\uparrow t$, then $t-s>0$. What we want is that if $t-s$ is arbitrarily small, so is $\|T(t)x-T(s)x\|\le \|T(t)-T(s)\| \|x\|$. But how does one bound $T(t)-T(s)$?
Assume the space is complete so that you can apply the Uniform Boundedness Principle. To prove boundedness of $\|T(t)\|$ in a neighborhood of $0$, assume the contrary. Then for each $n=1,2,3,\cdots$, there exists $t_n$ such that $\|T(t_n)\|\ge n$ and $t_n \downarrow 0$. By the Uniform Boundedness Principle, there exists $x \in X$ such that $\|T(t_n)x\|$ is unbounded, contrary to the assumption of right continuity of $T(t)x$ at $t = 0$. Hence $\|T(t)\|$ is bounded by some $M$ in some neighborhood of $0$, say $0 \le t \lt \delta$. From the exponential property, $\|T(t)\|\le M^{n}$ for $0 \le t < n\delta$.
Therefore, for fixed $t > 0$ and $0 \le s < t$ there is a constant $C$ such that \begin{align} \|T(s)x-T(t)x\|&=\|T(s)(x-T(t-s)x)\| \\ &\le C\|x-T(t-s)x\|\rightarrow 0 \mbox{ as } s\uparrow t. \end{align}
Reference: Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations