There are a few things you could do to simplify this.

• Kaplansky's theorem (Reference needed for: every idempotent in a C*-algebra is similar to a hermitian one) lets you show that every idempotent in a $C^*$-algebra is similar to a projection. Thus to prove your claim, you just need to show that every projection in $\mathcal{O}_n$ is a commutator element, because similarity preserves the property of being a commutator or linear combination of commutators.
• In $\mathcal{O}_n = C^*(s_1,\ldots,s_n)$ (here $s_1,\ldots,s_n$ are the usual generating isometries), the projections $s_i s_i^*$ are linear combinations of commutators. For any $i=1,\ldots,n$ you can take the isometry $s_i$ and then the commutator $[s_i^*,s_i]$ is $Q_i=\sum_{k \neq i} s_k s_k^*$. The set of all these projections $\{Q_i\}_{i =1}^n$ forms a basis for the vector space spanned by $\{s_1 s_1^*, \ldots, s_n s_n^*\}$. (This is just like showing the matrix with zeros on diagonal and $1$ everywhere else is invertible.)
• Every projection in $\mathcal{O}_n$ is in the same $K_0$-class as a projection of the form $s_1 s_1^* + \ldots +s_j s_j^*$ for some $j \leq n$. So all you need to do is show that belonging to the same $K_0$-class preserves being a linear combination of commutators. I think that the similarity characterization of deciding when two projections are in the same $K_0$-class works.

Sorry I realize that I haven't provided a definite answer, it just seems like all this would be quite cluttered as a comment.

There are a few things you could do to simplify this.

• Kaplansky's theorem (Reference needed for: every idempotent in a C*-algebra is similar to a hermitian one) lets you show that every idempotent in a $C^*$-algebra is similar to a projection. Thus to prove your claim, you just need to show that every projection in $\mathcal{O}_n$ is a linear combination of commutator elements, because similarity preserves the property of being a commutator or linear combination of commutators.
• In $\mathcal{O}_n = C^*(s_1,\ldots,s_n)$ (here $s_1,\ldots,s_n$ are the usual generating isometries), the projections $s_i s_i^*$ are linear combinations of commutators. For any $i=1,\ldots,n$ you can take the isometry $s_i$ and then the commutator $[s_i^*,s_i]$ is $Q_i=\sum_{k \neq i} s_k s_k^*$. The set of all these projections $\{Q_i\}_{i =1}^n$ forms a basis for the vector space spanned by $\{s_1 s_1^*, \ldots, s_n s_n^*\}$. (This is just like showing the matrix with zeros on diagonal and $1$ everywhere else is invertible.)
• Every projection in $\mathcal{O}_n$ is in the same $K_0$-class as a projection of the form $s_1 s_1^* + \ldots +s_j s_j^*$ for some $j \leq n$. So all you need to do is show that belonging to the same $K_0$-class preserves being a linear combination of commutators. I think that the similarity characterization of deciding when two projections are in the same $K_0$-class works.

Sorry I realize that I haven't provided a definite answer, it just seems like all this would be quite cluttered as a comment.

There are a few things you could do to simplify this.

• Kaplansky's theorem (Reference needed for: every idempotent in a C*-algebra is similar to a hermitian one) lets you show that every idempotent in a $C^*$-algebra is similar to a projection. Thus to prove your claim, you just need to show that every projection in $\mathcal{O}_n$ is a commutator element, because similarity preserves the property of being a commutator or linear combination of commutators.
• In $\mathcal{O}_n = C^*(s_1,\ldots,s_n)$ (here $s_1,\ldots,s_n$ are the usual generating isometries), the projections $s_i s_i^*$ are linear combinations of commutators. For any $i=1,\ldots,n$ you can take the isometry $s_i$ and then the commutator $[s_i,s_i^*]$ is $Q_i=\sum_{k \neq i} s_k s_k^*$. The set of all these projections $\{Q_i\}_{i =1}^n$ forms a basis for the vector space spanned by $\{s_1 s_1^*, \ldots, s_n s_n^*\}$. (This is just like showing the matrix with zeros on diagonal and $1$ everywhere else is invertible.)
• Every projection in $\mathcal{O}_n$ is in the same $K_0$-class as a projection of the form $s_1 s_1^* + \ldots +s_j s_j^*$ for some $j \leq n$. So all you need to do is show that belonging to the same $K_0$-class preserves being a linear combination of commutators. I think that the similarity characterization of deciding when two projections are in the same $K_0$-class works.

Sorry I realize that I haven't provided a definite answer, it just seems like all this would be quite cluttered as a comment.

There are a few things you could do to simplify this.

• Kaplansky's theorem (Reference needed for: every idempotent in a C*-algebra is similar to a hermitian one) lets you show that every idempotent in a $C^*$-algebra is similar to a projection. Thus to prove your claim, you just need to show that every projection in $\mathcal{O}_n$ is a commutator element, because similarity preserves the property of being a commutator or linear combination of commutators.
• In $\mathcal{O}_n = C^*(s_1,\ldots,s_n)$ (here $s_1,\ldots,s_n$ are the usual generating isometries), the projections $s_i s_i^*$ are linear combinations of commutators. For any $i=1,\ldots,n$ you can take the isometry $s_i$ and then the commutator $[s_i^*,s_i]$ is $Q_i=\sum_{k \neq i} s_k s_k^*$. The set of all these projections $\{Q_i\}_{i =1}^n$ forms a basis for the vector space spanned by $\{s_1 s_1^*, \ldots, s_n s_n^*\}$. (This is just like showing the matrix with zeros on diagonal and $1$ everywhere else is invertible.)
• Every projection in $\mathcal{O}_n$ is in the same $K_0$-class as a projection of the form $s_1 s_1^* + \ldots +s_j s_j^*$ for some $j \leq n$. So all you need to do is show that belonging to the same $K_0$-class preserves being a linear combination of commutators. I think that the similarity characterization of deciding when two projections are in the same $K_0$-class works.

Sorry I realize that I haven't provided a definite answer, it just seems like all this would be quite cluttered as a comment.