Banach algebra of smooth functions

I am not 100% sure that I am correctly interpreting your hypotheses, so let me write something which is more cumbersome but which hopefully applies to the situation you have in mind.

FACT 1. Let $A$ be a commutative Banach algebra and let $D:A\to A$ be a continuous derivation, i.e. a bounded linear map satisfying $D(ab)=aD(b)+D(a)b$ for all $a,b\in A$. Then every element in $D(A)$ is quasinilpotent, i.e. has spectral radius zero; equivalently, if we adjoin an identity to $A$ then $1+D(a)$ is invertible for all $a\in A$.

Proof: This is the Singer–Wermer theorem, whose proof can be found in various sources such as Bonsall and Duncan's book, and probably Rickart's, and almost certainly in the Big Book of Dales. I'll sketch the standard proof (a more elementary proofs was later given by Sinclair, IIRC).

The key observation is that the power series $\exp(D) :A \to A$ is a homomorphism, so replacing $D$ with $\lambda D$ for a complex scalar $\lambda$ one obtains a one-complex-parameter group of automorphisms of $A$. Composing with any fixed homomorphism $\varphi:A\to\mathbb C$, one obtains a family of homomorphisms $\psi_\lambda :=\varphi \circ \exp(\lambda D) : A\to \mathbb C$, and standard book-keeping shows that $\lambda\mapsto \psi_\lambda$ is holomorphic as a function $\mathbb C \to A^*$. But homomorphisms $A\to \mathbb C$ must all have norm $\leq 1$, so $\lambda\mapsto\psi_\lambda$ is a bounded entire function and hence constant by Liouville. By differentiating, one can deduce that for all $a\in A$ we have $\varphi(D(a))=0$; since this holds for all characters $\varphi$, Gelfand theory now tells us that $D(a)$ has spectral radius zero. QED.

REMARK 2. In fact, the previous result remains true without the assumption that the derivation $D$ is continuous. This is (much)$^{10}$ harder in my opinion and was a result of M. P. Thomas in Acta Mathematica in the 1980s.

COROLLARY 2. Equip $C^\infty({\mathbb T}^d)$ with its usual Fr'echet algebra structure, let $A$ be a commutative Banach algebra, and let $\Psi: A\to C^\infty({\mathbb T}^d)$ be an injective continuous homomorphism. Suppose that $\partial_j \Psi(A) \subseteq \Psi(A)$ for $j=1,\dots, d$ where $\partial_j$ is the partial derivative in the variable $\theta_j$. Then $\Psi(A)$ consists only of constant functions, and so $A$ is at most one-dimensional.

Proof. Let $D_j:A\to A$ be the unique linear map satisfying $\Psi\circ D_j = \partial_j \circ\Psi$. Since $\partial_j$ is continuous (for the Frechet algebra topology) and $\Psi$ is continuous, the closed graph theorem can be used to show that $D_j:A\to A$ is continuous. Clearly $D_j$ is also a derivation, since $\partial_j$ is and since $\Psi$ is an injective homomorphism.

Let $a\in A$. By Fact 1 (the Singer–Wermer theorem), $D_j(a)$ has spectral radius zero for each $j$, and hence $\partial_j\Psi(a)$ has spectral radius zero for each $j$. But the only element of $C^\infty({\mathbb T}^d)$ with spectral radius $0$ is the zero function. We conclude that $\Psi(a)$ has all partial derivatives equal to zero and by smoothness $\Psi(a)$ must be constant. QED.