Proof. First suppose that $A$ generates a strongly continuous contraction semigroup $\bracs{T_t|t \ge 0}$.
If the following integral does exist, then
\[\int_{0}^{\infty} e^{-t(\lambda - A)}dt = (\lambda - A)^{-1}\]
which suggests that
\[R_{\lambda} = \int_{0}^{\infty} e^{-\lambda t}T_{t}dt\]
is the inverse of $(\lambda - A)$. Since $t \mapsto T_{t}$ is continuous and $\norm{T_t}_{L(E; E)}\le 1$ for all $t \ge 0$, the integral converges absolutely as a Riemann integral with $\norm{R_\lambda}_{L(E; E)}\le \lambda^{-1}$.
For any $x \in E$ and $h > 0$,
\begin{align*}\frac{T_{h}R_{\lambda} x - R_{\lambda} x}{h}&= \int_{0}^{\infty} e^{-\lambda t}\frac{T_{h}T_{t}x - T_{t}x}{h}dt \\&= \frac{e^{\lambda h}}{h}\int_{h}^{\infty} e^{-\lambda s}(T_{s} - T_{s - h})xds \\&= \frac{e^{\lambda h}}{h}\braks{R_\lambda x - \int_0^h e^{-\lambda s}T_sxds}\\&- \frac{e^{\lambda h}}{h}\int_{h}^{\infty} e^{-\lambda s}T_{s - h}xds \\&= \frac{e^{\lambda h}}{h}\braks{R_\lambda x - \int_0^h e^{-\lambda s}T_sxds}\\&- \frac{1}{h}\int_{0}^{\infty} e^{-\lambda s}T_{s}xds \\&= \frac{e^{\lambda h}- 1}{h}R_{\lambda} x - \frac{e^{\lambda h}}{h}\int_{0}^{h}e^{-\lambda s}T_{s}xds\end{align*}
By the Fundamental Theorem of Calculus,
\[\lim_{h \downto 0}\frac{e^{\lambda h}}{h}\int_{0}^{h}e^{-\lambda s}= x\]
Therefore
\begin{align*}AR_{\lambda} x&= \lambda R_{\lambda} x - x \\ (\lambda - A)R_{\lambda} x&= x \\ (\lambda - A)R_{\lambda}&= \text{Id}\end{align*}
On the other hand, since $A$ is closed,
\[AR_{\lambda} x = \int_{0}^{\infty} e^{-\lambda t}T_{t}Axdt = R_{\lambda} Ax\]
Hence $R_{\lambda}(\lambda - A) = \text{Id}$ as well, so $R_{\lambda} = (\lambda - A)^{-1}$.
Now suppose that $A$ is a densely defined closed operator satisfying assumptions (1) and (2). For each $\lambda > 0$, let
\[A_{\lambda} = -\lambda + \lambda^{2}(\lambda - A)^{-1}\]
then for any $x \in D(A)$,
\begin{align*}A_{\lambda} x&= -\lambda(\lambda-A)(\lambda-A)^{-1}x + \lambda^{2}(\lambda-A)^{-1}x \\&= \lambda A(\lambda-A)^{-1}x = \lambda(\lambda-A)^{-1}Ax\end{align*}
By assumption (2), for any $y \in E$,
\begin{align*}\lambda(\lambda - A)^{-1}y - y&= \lambda(\lambda - A)^{-1}(y - \lambda^{-1}(\lambda-A)y) \\&= (\lambda - A)^{-1}(\lambda y - (\lambda-A)y) \\&= (\lambda-A)^{-1}Ay \\ \lim_{\lambda \to \infty}\lambda(\lambda - A)^{-1}y - y&= 0\end{align*}
Therefore for any $x \in D(A)$,
\[\limv{\lambda}A_{n}x = x\]
For each $\lambda > 0$, define
\[T^{\lambda}_{t} = \sum_{k = 0}^{\infty} \frac{t^{k}A_{\lambda}^{k}}{k!}\]
then $\bracsn{T^\lambda_t|t \ge 0}$ is a (very) strongly continuous semigroup with
\begin{align*}T_{t}^{\lambda}&= e^{-\lambda t}\sum_{k = 0}^{\infty} \frac{\lambda^{2k}t^{k}(\lambda - A)^{-k}}{k!}\\ \normn{T_t^\lambda}_{L(E; E)}&\le e^{-\lambda t}\sum_{k = 0}^{\infty} \frac{\lambda^{k} t^{k}}{k!}\le 1\end{align*}
so $\bracsn{T^\lambda_t|t \ge 0}$ is a contraction semigroup.
Now, for any $x \in D(A)$, $t \ge 0$, and $\lambda, \mu > 0$, by the Fundamental Theorem of Calculus,
\begin{align*}{(T_t^\lambda - T_t^\mu) x}&= \int_{0}^{t} \frac{d}{dx}(T_{s}^{\lambda} T_{t - s}^{\mu} x)ds \\&= \int_{0}^{t} T_{s}^{\lambda} T_{t - s}^{\mu} (A_{\lambda} - A_{\mu})xds \\ \normn{(T_t^\lambda - T_t^\mu) x}_{E}&\le t\norm{(A_\lambda - A_\mu)x}_{E}\end{align*}
This allows defining
\[T_{t}x = \limv{n}T_{t}^{n}x\]
for all $x \in D(A)$, and extend to $E$ by density. By continuity, $\bracs{T_t|t \ge 0}$ is a strongly continuous contraction semigroup.
Finally, let $B$ be the generator of $\bracs{T_t|t \ge 0}$, then for any $t \ge 0$, $x \in D(A)$, and $\lambda > 0$, by the Dominated Convergence Theorem,
\begin{align*}T_{t}^{\lambda} x&= x + \int_{0}^{t} T_{s}^{\lambda} Axds \\ T_{t}x&= x + \int_{0}^{t}T_{s} Axds\end{align*}
Hence $B$ is an extension of $A$. On the other hand, $(\lambda - A)$ and $(\lambda - B)$ are both bijective. Therefore $A$ is an extension of $B$, and $A= B$.$\square$