In fact, even a snake does not have to look beyond its nose to find a nice example (I post it here such that the original question does not get to long, discouraging potential readers).
Take $X$ as the complex-valued sequences with the usual vector-space structure and with norm
\begin{equation}
\left\|x\right\|_X^2=|x_1|^2+|x_2|^2+|x_3|^3...
\end{equation}
(i.e. $X=l^2(\mathbb{N})$)
and $Y$ again those sequences with the more relaxed norm
\begin{equation}
\left\|x\right\|_Y^2=\frac{1}{2}(|x_1|^2+\frac{1}{2!}|x_2|^2+\frac{1}{3!}|x_3|^2+...)
\end{equation}
For the role of closed operator $A:D(A)\subset Y \to Y$ we take the infinite diagonal matrix
\begin{equation}
A_{kl}=\left(k+i\sqrt{2k!-k^2}\right)\delta_{kl}.
\end{equation}
This seemingly "arbitrary" operator satisfies ($\forall x \in X$)
\begin{equation}
\left\|x\right\|_X=\left\|Ax\right\|_Y.
\end{equation}
which establishes that $D(A)=X$ (as a subset of $Y$ of course).
Next, let $Z\subset X$ be the sequences with finite support. $Z$ is dense in $X$ and $Y$ for their respective norms. It is straightforward to show that the sequence
\begin{equation}
u=(u_1,...,u_n,0,0...)
\end{equation}
is equal to $u(0)$ in the following solution of $u'=Au$ which is smooth in both the $X$ and $Y$-norm and unique for this initial condition:
\begin{equation}
u(t)=(u_1\exp(A_{11} t),...,u_n \exp(A_{kk}t),0,0...) (\in Z)
\end{equation}
It is not difficult to see that there is no uniform bound ($\forall u \in Y$ or $\in X$) to the quantities $\frac{\left\|u(t)\right\|_Y}{\left\|u\right\|_Y}$ or $\frac{\left\|u(t)\right\|_X}{\left\|u\right\|_X}$.
Assume however that $\left\|u\right\|_X=1$ which means in particular $|u_k|\leq 1$ $\forall k >0$. Then
\begin{equation}
\left\|u(t)\right\|_Y = \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k!}|u_k\exp(A_{kk}t)|^2 \leq \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k!}\exp(kt)^2=\frac{1}{2}\exp(\exp(2t))
\end{equation}
So far, I've produced all prerequisites stated in my question above, which I will show to be enough to produce a bilateral semigroup. That is, I've come up with a $u'=Au$-system which satisfies...
Definition Let $(Y,\left\|\right\|_Y)$ be a Banach space, a closed linear operator $A:D(A)\subset Y \to Y$ is called a "Cauchy-operator" if the set $Z\subset D(A)$, where $u\in Z$ i.f.f. a unique solution to $u'(t)=Au(t)$ ($\forall t \geq 0$) exists with initial condition $u$, is dense in $Y$. From now we define semigroup operators $R(t):Z \to D(A)$ through $u(t)=R(t)u$.
We call $A$ a "regular Cauchy-operator" if there exists moreover a vectorspace $X$ with $D(A) \subset X \subset Y$ with an accompanying norm $\left\|\right\|_X$ such that $(X,\left\|\right\|_X)$ is a Banach space and such that $Z$ is also dense in $(X,\left\|\right\|_X)$ and such that $\sup_{u \in Z}\frac{\left\|R(t)u\right\|_Y}{\left\|u\right\|_X} < \infty$ $\forall t >0$. $A$ is called a "smooth Cauchy-operator" if for every $t^* \in [0,\infty)$, there exists $\epsilon>0$ such that
\begin{equation}
\sup_{u \in Z, t\in [t^*-\epsilon,t^*+\epsilon]}\frac{\left\|R(t)u\right\|_Y}{\left\|u\right\|_X} < \infty
\end{equation}
Theorem Let $A:D(A)\subset Y \to Y$ be a Cauchy operator. $Z$ is a vector space and the $R(t):Z \to D(A)$-operators are linear.
Their range is contained in $Z \subset D(A)$ and $\forall s \geq 0$. Redefining them as $R(t):Z\subset Y \to Z \subset Y$ we have
\begin{equation}
R(s+t)u=R(s)R(t)u.
\end{equation}
proof: exercise.
Theorem If $A:D(A)\subset Y \to Y$ is a regular Cauchy operator. Then the bounded operators $R(t):Z\subset X \to Z \subset Y$ can be continuously extended to bounded operators
\begin{equation}
S(t):X \to Y.
\end{equation}
If at some point $t^*\in [0,\infty)$ and for some $\epsilon>0$, we have that $R(t)$ is uniformly bounded on $[t^*-\epsilon,t^*+\epsilon]$, then for all $u\in X$ the map
\begin{equation}
\varphi_u:[0,\infty)\to Y: t \mapsto S(t)u
\end{equation}
is continuous at $t^*$. As a consequence these maps are continuous if $A$ is a smooth Cauchy operator.
In this case the operators $S(t)$ constitute a bilateral one-parameter semigroup $G=\left\{S(t)\right\}_{t\geq 0}$. Note: I've edited the definition of bilateral semigroup slightly.
proof The first part of the theorem is straightforward and left to the reader. Keep in mind that the continuity of the $\varphi_u$ maps implies uniform continuity: for all compacts $C \subset [0,\infty)$ we have
\begin{equation}
\sup_{t\in C} \left\|S(t)\right\| \leq 0.
\end{equation}
For the second part of the theorem, we first have to show that elements of the form $\int_a^b S(s)u\text{d}s=\int_0^{\infty} \chi_{[a,b]}S(s)u\text{d}s$ or $\int_0^{\infty} f(s)S(s)u\text{d}s$
(where $f:[0,\infty)\to \mathbb{C}$ is differentiable) are in $X$. Secondly, we have to show that "$S(t) \int S(s)... = \int S(s+t)...$". (Important note: all the relevant Bochner integrals are of course defined in the Banach space $Y$!)
In this proof I'll only treat the second case, while the first is similar.
Take $(u_n)_n \in Z$ such that $\left\|u_n-u\right\|_X \to 0$. By the boundedness of the operators $S(t)$, we have
\begin{equation}
\left\|\int_0^{\infty} f(s)S(s)u_n\text{d}s - \int_0^{\infty} f(s)S(s)u\text{d}s\right\|_Y \to 0
\end{equation}
(exercise)
Moreover, we have
\begin{equation}
\left\|A\int_0^{\infty} f(s)S(s)u_n\text{d}s + f(0)u+\int_0^{\infty} f'(s)S(s)u\text{d}s\right\|_Y = \left\|\int_0^{\infty} f(s)AS(s)u_n\text{d}s + f(0)u+\int_0^{\infty} f'(s)S(s)u\text{d}s\right\|_Y =\left\|\int_0^{\infty} f(s)(S(s)u_n)'\text{d}s + f(0)u+\int_0^{\infty} f'(s)S(s)u\text{d}s\right\|_Y=\left\|-f(0)u_n - \int_0^{\infty} f'(s)S(s)u_n\text{d}s+ f(0)u+\int_0^{\infty} f'(s)S(s)u\text{d}s\right\|_Y \to 0
\end{equation}
So the closedness of $A:D(A) \subset Y \to Y$ implies that $v:=\int_0^{\infty} f(s)S(s)u\text{d}s \in D(A)\subset X$ and $Av =-f(0)u-\int_0^{\infty} f'(s)S(s)u\text{d}s$. To finish, one takes the following steps:
*$\eta:[0,\infty)\to Y: t \mapsto \int_0^{\infty} f(s)S(s+t)u_n\text{d}s=\int_t^{\infty} f(s-t)S(s)u_n\text{d}s$ is differentiable and $\eta'(t)=A\eta(t)$. Hence $\eta(t) \in Z$ for all $t\geq 0$.
*But then
\begin{equation}
\nu:[0,\infty)\to Y: h \mapsto S(t-h)\int_0^{\infty} f(s)S(s+h)u_n\text{d}s = S(t-h)\eta(h)
\end{equation}
is also differentiable and $\nu'(h)=0$. But then
\begin{equation}
S(t)\int_0^{\infty} f(s)S(s)u_n\text{d}s = \nu(0)=\nu(t)=\int_0^{\infty} f(s)S(s+t)u_n\text{d}s.
\end{equation}