Isomorphism between two binary algebraic structures

I'm studying from the book A First Course In Abstract Algebra, 7 Ed. by John B. Fraleigh. And I found the following exercise regarding isomorphisms between binary algebraic structures.

Let $F$ be the set of all functions $f$ mapping $\mathbb{R}$ into $\mathbb{R}$ that have derivatives of all orders. Determine whether the given map $\phi$ is an isomorphism of the first binary structure with the second. If it is not an isomorphism, why not? Section 3. Problem 13. $\langle F,+ \rangle$ with $\langle F,+ \rangle$ where $\phi(f)(x)=\int_0^xf(t)dt$.

I have been thinking about the solution for some days, however, I am blocked and cannot come up with a way to solve it. When checking the solutions manual, this is the answer.

No, because $\phi(f) = x+1$ has no solution $f\in F$.

My questions are:

To see that it does not have a solution, try to find one $f$ that satisfies the relation given. If it existed then, we would have that $$\varphi(f) = x + 1 \iff \int_{0}^{x} f(t) dt = x + 1$$ By the Fundamental Theorem of Calculus, we have that $f(x) = 1$. However, this is only a necessary condition for $f$ to satisfy the relation given. We need to check if it is indeed a solution. Checking that, we have $$\int_{0}^{x} f(t) dt = \int_{0}^{x} dt = x \neq x + 1$$ Therefore, it does not exist an $f$ which satisfies the above relation.

By the definition of isomorphism, we need $\varphi$ to be surjective. However, since $x + 1$ is an element of $F$ and it does not exist any $f$ such that $\varphi(f) = x + 1$, $\varphi$ cannot be surjective, hence, it is not an isomorphism.

EDIT: As a matter of fact, the surjective condition is the only reason that stops us from telling that $\varphi$ is an isomorphism, but it is a monomorphism.

The problem is that $\phi(f)$ is always a function whose graph passes through the origin: $$ \phi(f)(0) = \int_0^0 f(t)\,\mathrm{d}t = 0. $$

Claim: The set $F_0 = \{f \in F \mid f(0) = 0\}$ is a subgroup of $F$, and the map $\phi: F \to F_0$ is an isomorphism.

The easiest way to see that it's a bijection is to note that $D = \frac{\mathrm{d}}{\mathrm{d}x}: F_0 \to F$ is its inverse. Notice that for any $f \in F$, $$ D\phi(f)(x) = \frac{\mathrm{d}}{\mathrm{d}x} \int_0^x f(t)\,\mathrm{d}t = f(x) $$ by the Fundamental Theorem of Calculus. In the other direction, $$ \phi D(f)(x) = \int_0^xf'(t)\,\mathrm{d}t = f(x) - f(0), $$ and this is exactly $f(x)$ since $f \in F_0$. (You should also check that $D$ respects the additive structure, but that's straightforward.)

Now this presents an odd situation: a group is isomorphic with a proper subgroup! This is only possible since it's infinite.