Proof involving strongly continuous semigroups. Let $ (T(t))_{t \geq 0} $ be a $ C_{0} $-semigroup on a Hilbert space $ X $ with an infinitesimal generator $ A $, and let $ \rho \in (0,1) $.
I want to prove that $ \displaystyle \sup_{t \geq 0} \| T(t) - I \| \leq \rho $ is equivalent to $ A = 0 $.
I have already proven that when the inequality holds, then the infinitesimal generator $ A $ satisfies
$$
\forall \lambda > 0: \quad
\left\| \lambda (\lambda I - A)^{-1} - I \right\| \leq \rho,
$$
and I want to use this new inequality to show that $ A = 0 $ and thus $ T(t) = I $.
Define $ T_{\lambda} \stackrel{\text{df}}{=} \lambda (\lambda I - A)^{-1} - I $. Then it is clear that $ T \in \mathcal{L}(X) $ and $ \| T_{\lambda} \| \leq \rho $. Hence,
$$
     \| T_{\lambda} \| \leq \rho
\iff 1 - \| T_{\lambda} \| \leq 1 - \rho
\iff (1 - \| T_{\lambda} \|)^{-1} \leq (1 - \rho)^{-1}.
$$
We also have that $ (1 - \| T_{\lambda} \|)^{-1} = \dfrac{1}{1 - \| \lambda (\lambda I - A)^{-1} - I \|} $.
How do I proceed from here?
 A: As the OP has mentioned, we have
$$
\forall \lambda \in \mathbb{R}_{> 0}: \quad
T_{\lambda} \stackrel{\text{df}}{=}
\lambda (\lambda I - A)^{-1} - I \in B(\mathcal{H}) \quad \text{and} \quad
\| T_{\lambda} \|_{B(\mathcal{H})} \leq \rho.
$$
This is a consequence of the following theorem, whose statement is found in Theorem $ 13.35 $ of Walter Rudin’s Functional Analysis:

Theorem. Let $ (T(t))_{t \geq 0} $ be a $ C_{0} $-semigroup on a Hilbert space $ \mathcal{H} $. Denote its infinitesimal generator by $ A $. If $ C \in \Bbb{R}_{> 0} $ and $ \gamma \in \Bbb{R} $ are constants such that $ \| T(t) \|_{B(\mathcal{H})} \leq C e^{\gamma t} $ for each $ t \in \Bbb{R}_{\geq 0} $, then for any $ \lambda \in \Bbb{C} $ satisfying $ \Re(\lambda) > \gamma $, the map $ R_{\lambda}: \mathcal{H} \to \mathcal{H} $ defined by
  $$
\forall x \in \mathcal{H}: \quad
{R_{\lambda}}(x) \stackrel{\text{df}}{=}
\int_{0}^{\infty} e^{- \lambda t} \cdot [T(t)](x) ~ \mathrm{d}{t}
$$
  is a well-defined bounded linear operator (called the resolvent of $ (T(t))_{t \geq 0} $ at $ \lambda $) that has the following two properties:
  
  
*
  
*Its range is $ \text{Dom}(A) $.
  
*It inverts $ \lambda I - A $.
  

The given assumption on our $ C_{0} $-semigroup $ (T(t))_{t \geq 0} $ is that
$$
\forall t \in \Bbb{R}_{\geq 0}: \quad
\| T(t) - I \|_{B(\mathcal{H})} \leq \rho < 1.
$$
Hence, if we choose $ C = 1 + \rho $ and $ \gamma = 0 $, then
$$
\forall t \in \Bbb{R}_{\geq 0}: \quad
\| T(t) \|_{B(\mathcal{H})} \leq C e^{\gamma t}.
$$
Fix $ \lambda \in \Bbb{R}_{> 0} $ for the present moment. By the theorem, we have
$$
(\heartsuit) \qquad
\forall x \in \mathcal{H}: \quad
  {(\lambda I - A)^{-1}}(x)
= \int_{0}^{\infty} e^{- \lambda t} \cdot [T(t)](x) ~ \mathrm{d}{t}.
$$
Using the fact that $ \displaystyle \int_{0}^{\infty} \lambda e^{- \lambda t} ~ \mathrm{d}{t} = 1 $, we also have
$$
(\diamondsuit) \qquad
\forall x \in \mathcal{H}: \quad
x = \int_{0}^{\infty} \lambda e^{- \lambda t} \cdot x ~ \mathrm{d}{t}.
$$
Multiplying $ (\heartsuit) $ by $ \lambda $ and then subtracting $ (\diamondsuit) $ from that yields
\begin{align}
\forall x \in \mathcal{H}: \quad
    {T_{\lambda}}(x)
& = \left( \lambda (\lambda I - A)^{-1} - I \right) \! (x) \\
& = \int_{0}^{\infty} \lambda e^{- \lambda t} \cdot [T(t) - I](x) ~ \mathrm{d}{t}.
\end{align}
We thus get
\begin{align}
\forall x \in \mathcal{H}: \quad
       \| {T_{\lambda}}(x) \|_{\mathcal{H}}
& \leq \int_{0}^{\infty}
       \left\|
       \lambda e^{- \lambda t} \cdot [T(t) - I](x)
       \right\|_{B(\mathcal{H})} ~ \mathrm{d}{t} \\
& \leq \int_{0}^{\infty}
       \lambda e^{- \lambda t} \cdot
       \| T(t) - I \|_{B(\mathcal{H})} \cdot \| x \|_{\mathcal{H}} ~
       \mathrm{d}{t} \\
& \leq \left( \int_{0}^{\infty} \lambda e^{- \lambda t} ~ \mathrm{d}{t} \right)
       \rho \| x \|_{\mathcal{H}} \\
& =    \rho \| x \|_{\mathcal{H}}.
\end{align}
Therefore, $ \| T_{\lambda} \|_{B(\mathcal{H})} \leq \rho $, as desired.

Claim: $ \text{Dom}(A) = \mathcal{H} $ and $ A \in B(\mathcal{H}) $.

Proof of Claim
As $ \| T_{\lambda} \| \leq \rho < 1 $, a standard result about Banach algebras tells us that $ I + T_{\lambda} $ is an invertible element of $ B(\mathcal{H}) $ whose inverse is the ‘geometric series’
$$
\sum_{k = 0}^{\infty} (-1)^{k} \cdot (T_{\lambda})^{k}.
$$
As $ I + T_{\lambda} = \lambda (\lambda I - A)^{-1} $, we obtain
\begin{align}
    \text{Dom}(A)
& = \text{Range} \! \left( (\lambda I - A)^{-1} \right) \\
& = \text{Range} \! \left( \lambda (\lambda I - A)^{-1} \right) \\
& = \text{Range}(I + T_{\lambda}) \\
& = \mathcal{H}. \qquad
    (\text{As $ I + T_{\lambda} $ is invertible, hence surjective.})
\end{align}
As $ A $ is a closed operator (this is a standard fact in the theory of $ C_{0} $-semigroups), it follows that $ A \in B(\mathcal{H}) $. (Note: By the Closed Graph Theorem, a closed operator that is defined everywhere is bounded.) $ \quad \blacksquare $
Observe now that
$$
  \frac{1}{\lambda} (\lambda I - A)
= (I + T_{\lambda})^{-1}
= \sum_{k = 0}^{\infty} (-1)^{k} \cdot (T_{\lambda})^{k},
$$
so
$$
     \left\| \frac{1}{\lambda} (\lambda I - A) \right\|_{B(\mathcal{H})}
\leq \sum_{k = 0}^{\infty} \| T_{\lambda} \|_{B(\mathcal{H})}^{k}
\leq \sum_{k = 0}^{\infty} \rho^{k}
=    \frac{1}{1 - \rho}.
$$
It follows readily that
$$
(\spadesuit) \qquad
\| \lambda I - A \|_{B(\mathcal{H})} \leq \frac{\lambda}{1 - \rho}.
$$
Let $ \lambda \to 0^{+} $. The left-hand side of $ (\spadesuit) $ goes to $ \| A \|_{B(\mathcal{H})} $, while the right-hand side goes to $ 0 $. By the Squeeze Theorem, $ \| A \|_{B(\mathcal{H})} = 0 $, so we arrive at the desired conclusion that $ A = 0_{B(\mathcal{H})} $.
