Here's a positive result towards this answer. This doesn't feel like something that could be adapted to a general proof, but it at least is an easy result backing up the result quoted in Halmos' book.

Consider a unitary, diagonalisable operator, U, on a separable Hilbert space, $H$. So there is some countable orthonormal basis on which $U$ is diagonal. Index it by the **integers** and let $S$ be the shift operator (which is unitary). Let $\lambda_i$ be the corresponding eigenvalues of $U$. For each $i$, let $\mu_i$ be defined by $\mu_0 = 1$ and $\mu_{i+1}/\mu_i = \lambda_i$. Let $V$ be the operator which acts on $e_i$ by $\mu_i$. As the $\lambda_i$ were all unitary, $V$ is a unitary operator. Then

$$
V^{-1}S^{-1}VSe^i = V^{-1}S^{-1}Ve_{i+1} = \mu_{i+1}V_{-1}S^{-1}e_{i+1} = \mu_{i+1}V^{-1}e_i = \mu_{i+1}/\mu_i e_i = \lambda_ie_i = Ue_i
$$

So $V^{-1}S^{-1}VS = U$.

In particular, one can get $\lambda I$ this way for $\lambda \in S^1$.

As I said, this feels too specific to adapt to a general proof, but it is a simple example.

Edit: In a similar way, we can get $\lambda I$ for any non-zero $\lambda$. Let's stay with separable Hilbert spaces for simplicity. To make things clear, let $H$ be the Hilbert space we're interested in, and let $H'$ be a standard reference Hilbert space. Let $H_0 \subseteq H$ be a closed subspace of infinite dimension and codimension. Split the complement as a sum $H_1 \oplus H_2 \oplus \dots$ where each $H_i$ is an infinite dimensional closed subspace. So $H \cong \bigoplus H_j$ and each $H_j$ are isomorphic to each other and to $H'$. Choose $A \in GL(H')$. Let $S_1$ be the switch operator that swaps $H_{2i}$ with $H_{2i+1}$. Let $T_1 = (A,I,A,I,\dots)$. Then $T_1 S_1 T_1^{-1} S_1^{-1}$ is $(A,A^{-1},A,A^{-1},\dots)$. Now let $S_2$ be the switch operator that swaps $H_{2i-1}$ with $H_{2i}$ and leaves $H_0$ alone. Let $T_2 = (I,A,I,A,\dots)$. Then $T_2 S_2 T_2^{-1} S_2^{-1} = (I,A,A^{-1},A,A^{-1},\dots)$. So multiplying these together, we get $(A,I,I,I,\dots)$.

Thus for any splitting $H \cong H_0 \oplus H_0^\perp$, and $A \in GL(H_0)$, $B \in GL(H_0^\perp)$ we can get

$$
\begin{bmatrix} A & 0 \\\\ 0 & B \end{bmatrix}.
$$

In particular, $\lambda I$ for any $\lambda \in \mathbb{C}^*$.