Let $T$ be a $C_0$-semigroup on the Banach space $E$. Then $\exists$ $M\ge 0$ and $\omega\in\mathbb{R}$ such that
a) \begin{equation*} \|T(t)\|\le M e^{\omega t}\quad\forall\quad t\ge 0 \end{equation*} b) For all $x\in E$, the function $[0,\infty)\ni t\mapsto T(t)x\in E$ is continuous. In other words, the function $T$ is strongly continuous.
c) If $T$ is a $C_0$-group on $E$, then $\exists\,M\ge 0$ and $\omega\in\mathbb{R}$ such that \begin{equation*} \|T(t)\|\le M e^{\omega|t|}\quad\forall\quad t\in\mathbb{R} \end{equation*} For all $x\in E$, the function $\mathbb{R}\ni t\mapsto T(t)x\in E$ is continuous. In other words, the function $T$ is strongly continuous.
I am fine with (a), (b), (c), except the last part of (c), i.e. to prove $T$ is strongly continuous on the whole $\mathbb{R}$. I disregard posting my attempts of (a) to save space. So let me start with the solution of (b)
To show strong continuity of $T$ on $[0,\infty]$ one needs to show the map $t\mapsto T(t)x$ is continuous $\forall\,x\in E$. So, let $x\in X$, $t>0$. Then $T(t+h)x-T(t)x=T(t)(T(h)x-x)\to 0$ as $h\to 0^{+}$, then \begin{equation*} \|T(t+h)x-T(t)x\|\le \left(\sup_{0\le s\le t}\|T(s)\|\right)(T(h)x-x)\to 0 \end{equation*} as $h\to 0^+$. This proves the right-sided continuity of $T(\cdot)x$ of $T(\cdot)x$. In order to prove the left-sided continuity, we let $-t\le h<0$ and write $T(t+h)x-T(t)x=T(t+h)(x-T(-h)x).$ Then one obtains \begin{equation*} \|T(t+h)x-T(t)x\|\le \left(\sup_{0\le s\le t}\|T(s)\|\right)(x-T(-h)x)\to 0 \end{equation*} as $h\to 0^-$.
(c) Here we are proving strong continuity for $T$ on the whole $\mathbb{R}$. So here, we need to show the orbit $T(\cdot)x$ is continuous for any given $x\in X$. In (b) we already proved that the restriction of $T$ to $[0,\infty)$ is a $C_0$-semigroup and it follows from $(b)$ that $T(\cdot)x$ is continuous on $[0,\infty)$. Let $t\le 0$. Then $T(t+h)x-T(t)x=T(t-1)(T(1+h)x-T(1)x)\to 0$ ($h\to 0$), and this implies that $T(\cdot)x$ is continuous on $\mathbb{R}$. As a consequence, the function $[0,\infty)\ni t\mapsto T(-t)\in\mathcal{L}(X)$ is a $C_0$ semigroup, and hence satisfies an estimate as in (a). Putting the estimates for the $C_0$-semigroups $t\mapsto T(t)$ and $t\mapsto T(-t)$ together one obtains the asserted estimate.
For (c), the solution is given. But I cannot understand the logic at all, especially "Let $t\le 0$. Then $T(t+h)x-T(t)x=T(t-1)(T(1+h)x-T(1)x)\to 0$ ($h\to 0$), and this implies that $T(\cdot)x$ is continuous on $\mathbb{R}$." What is going on here compare the $t>0$ case ? Why factor out $T(t-1)$, why analysing the continuity as $h\to 0$ as opposed what has been done for half line case (looking at $h\to 0^+$ and $h\to 0^-$