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Why make $A$ an algebra? Integration is fundamentally a linear operation that has at best a complicated relationship with multiplication on $C^\infty(\mathbb{R})$. And I see that Scott Carnahan has just made the same point in another answer... Scott's the kernel of what I was going to suggest as well: asymptotic growth classes. Let me expand on that.

Let $I_+ \subset C^\infty(\mathbb{R})$ be the ideal of functions that vanish on some neighborhood of $+\infty$ (for definiteness, say, on at least one interval of the form $[a,+\infty)$ with $a\in\mathbb{R}$). Let $A_+ = C^\infty(\mathbb{R})/I_+$. The quotient $A_+$ is an $\mathbb{R}$-algebra whose elements capture the rates of asymptotic growth at $+\infty$. Let $1_+$ be the image of the constant function $1\in C^\infty(\mathbb{R})$ under the quotient map. Define $I_-$, $A_-$ and $1_0$ 1_-$ in the same way, by replacing $+\infty$ with $-\infty$.

Now, let $B=A_+\oplus A_-$ and $N\subset B$ be the linear subspace spanned by the element $1_+\oplus 1_-$. And finally let $A = B/N$, where we are just taking the quotient of linear spaces (the ring property of $B$ ceases to be of importance).

Any smooth function $f\in C^\infty(\mathbb{R})$ has an indefinite integral $f_a(x) = \int_a^x f(y) dy$ that is also in $C^\infty(\mathbb{R})$. Applying the quotient maps above, we end up with an image $[f_a]$ of $f_a$ in the linear space $A$. The fact that the quotient images of constant functions give zero shows that the images indefinite integrals with different base points (say $f_a$ and $f_b$) coincide. I think that letting $T(f) = [f_a]$ will satisfy all the properties that you wanted of an "integration" map.
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Why make $A$ an algebra? Integration is fundamentally a linear operation that has at best a complicated relationship with multiplication on $C^\infty(\mathbb{R})$. And I see that Scott Carnahan has just made the same point in another answer... Scott's the kernel of what I was going to suggest as well: asymptotic growth classes. Let me expand on that.

Let $I_+ \subset C^\infty(\mathbb{R})$ be the ideal of functions that vanish on every some neighborhood of $+\infty$ (for definiteness, say, on at least one interval of the form $[a,+\infty)$ with $a\in\mathbb{R}$). Let $A_+ = C^\infty(\mathbb{R})/I_+$. The quotient $A_+$ is an $\mathbb{R}$-algebra whose elements capture the rates of asymptotic growth at $+\infty$. Let $1_+$ be the image of the constant function $1\in C^\infty(\mathbb{R})$ under the quotient map. Define $I_-$, $A_-$ and $1_0$ in the same way, by replacing $+\infty$ with $-\infty$.

Now, let $B=A_+\oplus A_-$ and $N\subset B$ be the linear subspace spanned by the element $1_+\oplus 1_-$. And finally let $A = B/N$, where we are just taking the quotient of linear spaces (the ring property of $B$ ceases to be of importance).

Any smooth function $f\in C^\infty(\mathbb{R})$ has an indefinite integral $f_a(x) = \int_a^x f(y) dy$ that is also in $C^\infty(\mathbb{R})$. Applying the quotient maps above, we end up with an image $[f_a]$ of $f_a$ in the linear space $A$. The fact that the quotient images of constant functions give zero shows that the images indefinite integrals with different base points (say $f_a$ and $f_b$) coincide. I think that letting $T(f) = [f_a]$ will satisfy all the properties that you wanted of an "integration" map.
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Why make $A$ an algebra? Integration is fundamentally a linear operation that has at best a complicated relationship with multiplication on $C^\infty(\mathbb{R})$. And I see that Scott Carnahan has just made the same point in another answer... Scott's the kernel of what I was going to suggest as well: asymptotic growth classes. Let me expand on that.

Let $I_+ \subset C^\infty(\mathbb{R})$ be the ideal of functions that vanish on every neighborhood of $+\infty$ (for definiteness, say, on at least one interval of the form $[a,+\infty)$ with $a\in\mathbb{R}$). Let $A_+ = C^\infty(\mathbb{R})/I_+$. The quotient $A_+$ is an $\mathbb{R}$-algebra whose elements capture the rates of asymptotic growth at $+\infty$. Let $1_+$ be the image of the constant function $1\in C^\infty(\mathbb{R})$ under the quotient map. Define $I_-$, $A_-$ and $1_0$ in the same way, by replacing $+\infty$ with $-\infty$.

Now, let $B=A_+\oplus A_-$ and $N\subset B$ be the linear subspace spanned by the element $1_+\oplus 1_-$. And finally let $A = B/N$, where we are just taking the quotient of linear spaces (the ring property of $B$ ceases to be of importance).

Any smooth function $f\in C^\infty(\mathbb{R})$ has an indefinite integral $f_a(x) = \int_a^x f(y) dy$ that is also in $C^\infty(\mathbb{R})$. Applying the quotient maps above, we end up with an image $[f_a]$ of $f_a$ in the linear space $A$. The fact that the quotient images of constant functions give zero shows that the images indefinite integrals with different base points (say $f_a$ and $f_b$) coincide. I think that letting $T(f) = [f_a]$ will satisfy all the properties that you wanted of an "integration" map.