LSpice asked me to expand my old comment into an answer.

Let me write $R$ for whatever base commutative ring you are working over, $V$ for your $R$-module, and $TV := T_R V$ for its "tensor algebra", defined by: for any (not necessarily commutative) $R$-algebra $A$, maps $T_R V \to A$ of $R$-algebras are the same as maps $V \to A$ of $R$-modules. In other words, $T_R$ is the left adjunct to the forgetful map $\{R$-algebras$\} \to \{R$-modules$\}$. Since $T_R$ is a functor, and since the group $R^\times$ of invertible elements of $R$ acts by rescaling on $V$, we get automatically a natural action of $R^\times$ on $T_R V$.

Now for any commutative $R$-algebra $S$, I claim that there is a natural equality $$ TV \otimes S := T_S (V \otimes_R S) \cong (T_R V) \otimes_R S.$$ Indeed, to show this I simply need to show that the RHS satisfies the correct universal property for the LHS, i.e. test it against maps into $S$-algebras $A$. But if $B$ is any $R$-algebra, and $A$ is any $S$-algebra, then the hom-tensor adjunction identifies $R$-algebra maps $B \to A$ with $S$-algebra map $B \otimes_R S \to A$. The claim follows.

But $S^\times$ acts on $TV \otimes S$, extending the $R^\times$-action from before. These actions compile into an action on $TV$ by the affine algebraic group $\mathbb{G}_m = \mathrm{Spec}_R(R[z^{\pm 1}]) : S \mapsto S^\times$.

Now, I claim that for any $R$-module $W$, an action on $W$ by $\mathbb{G}_m$ is the same data as a $\mathbb{Z}$-grading of $W$. But this is just the Peter–Weyl theorem and the fact that the regular representation $\mathcal{O}(\mathbb{G}_m) = R[z^{\pm 1}]$ breaks up as a direct sum of $\mathbb{Z}$-many one-dimensional modules.

In detail, because we are working with algebraic groups, it is well-defined to talk about "the subspace $W_n$ of $W$ on which $z \in \mathbb{G}_m$ acts by $z^n$", and it is easy to show that this is a direct summand. The only slightly non-easy fact is that the inclusion $\bigoplus_n W_n \subset W$ is an isomorphism, or in other words that every element of $W$ is a finite sum of elements with well-defined eigenvalues. The cleanest way to show this is to convince yourself that an action of an affine algebraic group $G$ on a module $W$ consists of a coaction $W \to \mathcal{O}(G) \otimes W$; then the statement that we need (about decomposing as a finite sum) follows just the fact that elements of any tensor product, and in particular $\mathcal{O}(G) \otimes W$, are finite sums of pure tensors.

LSpice asked me to expand my old comment into an answer.

Let me write $R$ for whatever base commutative ring you are working over, $V$ for your $R$-module, and $TV := T_R V$ for its "tensor algebra", defined by: for any (not necessarily commutative) $R$-algebra $A$, maps $T_R V \to A$ of $R$-algebras are the same as maps $V \to A$ of $R$-modules. In other words, $T_R$ is the left adjunct to the forgetful map $\{R$-algebras$\} \to \{R$-modules$\}$. Since $T_R$ is a functor, and since the group $R^\times$ of invertible elements of $R$ acts by rescaling on $V$, we get automatically a natural action of $R^\times$ on $T_R V$.

Now for any commutative $R$-algebra $S$, I claim that there is a natural equality $$ TV \otimes S := T_S (V \otimes_R S) \cong (T_R V) \otimes_R S.$$ Indeed, to show this I simply need to show that the RHS satisfies the correct universal property for the LHS, i.e. test it against maps into $S$-algebras $A$. But if $B$ is any $R$-algebra, and $A$ is any $S$-algebra, then the hom-tensor adjunction identifies $R$-algebra maps $B \to A$ with $S$-algebra map $B \otimes_R S \to A$. The claim follows.

But $S^\times$ acts on $TV \otimes S$, extending the $R^\times$-action from before. These actions compile into an action on $TV$ by the affine algebraic group $\mathbb{G}_m = \mathrm{Spec}_R(R[z^{\pm 1}]) : S \mapsto S^\times$.

Now, I claim that for any $R$-module $W$, an action on $W$ by $\mathbb{G}_m$ is the same data as a $\mathbb{Z}$-grading of $W$. But this is just the Peter–Weyl theorem and the fact that the regular representation $\mathcal{O}(\mathbb{G}_m) = R[z^{\pm 1}]$ breaks up as a direct sum of $\mathbb{Z}$-many one-dimensional modules.

In detail, because we are working with algebraic groups, it is well-defined to talk about "the subspace $W_n$ of $W$ on which $z \in \mathbb{G}_m$ acts by $z^n$", and it is easy to show that this is a direct summand. The only slightly non-easy fact is that the inclusion $\bigoplus_n W_n \subset W$ is an isomorphism, or in other words that every element of $W$ is a finite sum of elements with well-defined eigenvalues. The cleanest way to show this is to convince yourself that an action of an affine algebraic group $G$ on a module $W$ consists of a coaction $W \to \mathcal{O}(G) \otimes W$; then the statement that we need (about decomposing as a finite sum) follows just the fact that elements of any tensor product, and in particular $\mathcal{O}(G) \otimes W$, are finite sums of pure tensors.