By adapting Popa's short (and slick) proof in

Popa, Sorin, On commutators in properly infinite W*-algebras, Invariant subspaces and other topics, 6th int. Conf., Timisoara and Herculane/Rom. 1981, Operator Theory: Adv. Appl. 6, 195-207 (1982). ZBL0529.46043.

of the Wiener-Weilandt theorem that the identity is not the commutator of two bounded operators, one can directly answer the OP's question without the need to invoke the result of Kadison, as follows.

By multiplying $D$ by a scalar if necessary we may assume that the operator norm of the derivation $D$ is at most one: $\|Dx\| \leq \|x\|$ for all $x \in A$. If for contradiction we had $Dx=1$ for some $x \in A$, then we have $D(x^n) = n x^{n-1}$ for all $n \geq 1$, hence on taking norms we have $$ n \|x^{n-1}\| = \|D(x^n)\| \leq \| x^n \|.$$ Iterating this we conclude that $\|x^n\| \geq n!$. But $\|x^n\|$ can grow at most exponentially in $n$, giving the required contradiction.

This argument in fact gives the more quantitative lower bound $\|Dx-1\| \geq \exp( - \|D\|_{op} \|x\|)$ when worked out more carefully; see Popa's paper for further discussion. There are also constructions that give somewhat matching upper bounds; see this recent paper of myself (for the case $A=B(H)$) and of Krishna-Johnson (for more general C^* algebras).

By adapting Popa's short (and slick) proof in

Popa, Sorin, On commutators in properly infinite W*-algebras, Invariant subspaces and other topics, 6th int. Conf., Timisoara and Herculane/Rom. 1981, Operator Theory: Adv. Appl. 6, 195-207 (1982). ZBL0529.46043.

of the Wiener-Weilandt theorem that the identity is not the commutator of two bounded operators, one can directly answer the OP's question without the need to invoke the result of Kadison, as follows.

If for contradiction we had $Dx=1$ for some $x \in A$, then we have $D(x^n) = n x^{n-1}$ for all $n \geq 1$, hence $D^n(x^n)=n!$. But $\|D^n(x^n)\| \leq \|D\|_{op}^n \|x\|^n$ grows at most exponentially in $n$, giving the required contradiction.

This argument in fact gives the more quantitative lower bound $\|Dx-1\| \geq \exp( - \|D\|_{op} \|x\|)$ when worked out more carefully; see Popa's paper for further discussion. There are also constructions that give somewhat matching upper bounds; see this recent paper of myself (for the case $A=B(H)$) and of Krishna-Johnson (for more general C^* algebras).

By adapting Popa's proof in

Popa, Sorin, On commutators in properly infinite W*-algebras, Invariant subspaces and other topics, 6th int. Conf., Timisoara and Herculane/Rom. 1981, Operator Theory: Adv. Appl. 6, 195-207 (1982). ZBL0529.46043.

of the Wiener-Weilandt theorem that the identity is not the commutator of two bounded operators, one can directly answer the OP's question without the need to invoke the result of Kadison, as follows.

By multiplying $D$ by a scalar if necessary we may assume that the operator norm of the derivation $D$ is at most one: $\|Dx\| \leq \|x\|$ for all $x \in A$. If for contradiction we had $Dx=1$ for some $x \in A$, then we have $D(x^n) = n x^{n-1}$ for all $n \geq 1$, hence on taking norms we have $$ n \|x^{n-1}\| = \|D(x^n)\| \leq \| x^n \|.$$ Dividing by $n!$ and summing we conclude that $$ \sum_{n=0}^\infty \frac{\|x^n\|}{n!} \leq \sum_{n=1}^\infty \frac{\|x^n\|}{n!}.$$ Both sides are convergent, hence we may cancel to obtain $1 \leq 0$, which is absurd.

This argument in fact gives the more quantitative lower bound $\|Dx-1\| \geq \exp( - \|D\|_{op} \|x\|)$ when worked out more carefully; see Popa's paper for further discussion. There are also constructions that give somewhat matching upper bounds; see this recent paper of myself (for the case $A=B(H)$) and of Krishna-Johnson (for more general C^* algebras).

By adapting Popa's short (and slick) proof in

Popa, Sorin, On commutators in properly infinite W*-algebras, Invariant subspaces and other topics, 6th int. Conf., Timisoara and Herculane/Rom. 1981, Operator Theory: Adv. Appl. 6, 195-207 (1982). ZBL0529.46043.

of the Wiener-Weilandt theorem that the identity is not the commutator of two bounded operators, one can directly answer the OP's question without the need to invoke the result of Kadison, as follows.

By multiplying $D$ by a scalar if necessary we may assume that the operator norm of the derivation $D$ is at most one: $\|Dx\| \leq \|x\|$ for all $x \in A$. If for contradiction we had $Dx=1$ for some $x \in A$, then we have $D(x^n) = n x^{n-1}$ for all $n \geq 1$, hence on taking norms we have $$ n \|x^{n-1}\| = \|D(x^n)\| \leq \| x^n \|.$$ Iterating this we conclude that $\|x^n\| \geq n!$. But $\|x^n\|$ can grow at most exponentially in $n$, giving the required contradiction.

This argument in fact gives the more quantitative lower bound $\|Dx-1\| \geq \exp( - \|D\|_{op} \|x\|)$ when worked out more carefully; see Popa's paper for further discussion. There are also constructions that give somewhat matching upper bounds; see this recent paper of myself (for the case $A=B(H)$) and of Krishna-Johnson (for more general C^* algebras).