The ball formulation of the Kaplansky density theorem in nonselfadjoint algebras

Let $$H$$ be a separable Hilbert space. Let $$A \subseteq B(H)$$ be a finitely generated unital algebra. Let $$M$$ be the strong operator topology closure of $$A.$$ Let $$B_A$$ be the closed ball in $$A$$ and $$B_M$$ be the closed ball in $$M.$$ Is $$B_M$$ the strong operator topology closure of $$B_A?$$ (More likely, is there a counterexample?)

W. R. Wogen, Some counterexamples in nonselfadjoint operator algebras, Ann. Math. 126 (1987), 415-427, constructs an operator $$T$$ for which the WOT and weak* closures of the algebra generated by $$T$$ and $$I$$ are distinct. Since the WOT closure equals the SOT closure (and WOT and weak* agree on bounded sets), this gives a counterexample to your version.