2 Added remark pointing out that another example is possible using similar methods.

In what follows I show that such an operator exists if $E$ can be written (isometrically) as the $\ell_\infty$-direct sum of two (nonzero) subspaces (I have not tried the Hilbert space case, but I started writing my answer before the edits were made to the question.)

Let $E = X\oplus_\infty Y$, where $X$ and $Y$ are nonzero (infinite dimensional, if you like). Each $V\in L(E)$ satisfies $\Vert V \Vert = \max ( \Vert P_X V \Vert, \Vert P_Y V\Vert )$, where $P_X$ and $P_Y$ denote the projections onto the complemented subspaces $X$ and $Y$.

Let $T= P_X$ and $S=3P_Y$, so that $\Vert T-S\Vert =3=\Vert T+S\Vert$. To construct the desired example, we show that if $\Vert R-S\Vert <1$, then $\Vert T-R\Vert = \Vert T+R\Vert$. So take such $R$ and note that then $\Vert P_YR \Vert >2$ and $\Vert P_XR\Vert<1$. It follows that $$\Vert T-R\Vert = \max (\Vert P_X(T-R) \Vert ,\Vert P_Y(T-R)\Vert ) = \max (\Vert T-P_XR \Vert ,\Vert P_YR\Vert ) = \Vert P_Y R\Vert$$ (since $\Vert T-P_XR \Vert \leq \Vert T\Vert + \Vert P_XR \Vert$<2 and $\Vert P_Y R\Vert >2$).

Similarly, we conclude that $\Vert T+R\Vert = \Vert P_Y R\Vert$, hence $\Vert T+R\Vert = \Vert T-R\Vert$.

Edit: Note that since each $U\in L(X\oplus_1 Y)$ satisfies $\Vert U\Vert = \max (\Vert UP_X\Vert , \Vert UP_Y\Vert )$, a similar construction gives an example of such a ball for spaces isometrically isomorphic to $X\oplus_1 Y$ for nonzero $X$ and $Y$.

1

In what follows I show that such an operator exists if $E$ can be written (isometrically) as the $\ell_\infty$-direct sum of two (nonzero) subspaces (I have not tried the Hilbert space case, but I started writing my answer before the edits were made to the question.)

Let $E = X\oplus_\infty Y$, where $X$ and $Y$ are nonzero (infinite dimensional, if you like). Each $V\in L(E)$ satisfies $\Vert V \Vert = \max ( \Vert P_X V \Vert, \Vert P_Y V\Vert )$, where $P_X$ and $P_Y$ denote the projections onto the complemented subspaces $X$ and $Y$.

Let $T= P_X$ and $S=3P_Y$, so that $\Vert T-S\Vert =3=\Vert T+S\Vert$. To construct the desired example, we show that if $\Vert R-S\Vert <1$, then $\Vert T-R\Vert = \Vert T+R\Vert$. So take such $R$ and note that then $\Vert P_YR \Vert >2$ and $\Vert P_XR\Vert<1$. It follows that $$\Vert T-R\Vert = \max (\Vert P_X(T-R) \Vert ,\Vert P_Y(T-R)\Vert ) = \max (\Vert T-P_XR \Vert ,\Vert P_YR\Vert ) = \Vert P_Y R\Vert$$ (since $\Vert T-P_XR \Vert \leq \Vert T\Vert + \Vert P_XR \Vert$<2 and $\Vert P_Y R\Vert >2$).

Similarly, we conclude that $\Vert T+R\Vert = \Vert P_Y R\Vert$, hence $\Vert T+R\Vert = \Vert T-R\Vert$.