By the functional equation, it suffices to prove that $f$ is continuous at one point.
The fact that $f$ is of first Baire class is very straightforward:
$$
f(x) = \lim_{n \to \infty} \frac{F(x+1/n)-F(x)}{1/n}
$$
is a pointwise limit of continuous functions.
Now a function of first Baire class has a comeager $G_\delta$-set of points of continuity. Done.
Indeed, enumerate the open intervals with rational endpoints as $\langle I_n \mid n \in \omega\rangle$. Then
$$
f \text{ is discontinuous at }x \iff \exists n\in \omega : x \in f^{-1}[I_n] \setminus \operatorname{int}f^{-1}[I_n]
$$
Since $f$ is of first Baire class, $f^{-1}[I_n]$ is an $F_\sigma$ and so is $f^{-1}[I_n] \setminus \operatorname{int} f^{-1}[I_n]$. Therefore we can write
$$
f^{-1}[I_n] \setminus \operatorname{int} f^{-1}[I_n] = \bigcup_{k \in \omega} F_{k}^{n}
$$
for some sequence $\langle F_{k}^n \mid k \in \omega\rangle$ of closed sets. Observe that $F_{k}^n$ has no interior, so the set of points of discontinuity of $f$ is
$$
\bigcup_{n \in \omega} f^{-1}[I_n] \setminus \operatorname{int}f^{-1}[I_n] = \bigcup_{n\in\omega} \bigcup_{k\in\omega} F_{k}^n,
$$
a countable union of closed and nowhere dense sets.
If $0$ is in $\mathbb{N}$ with the convention $\varphi(0)=0$, then there is a unique solution $f(n)=0$ for every $n\in\mathbb{N}$. From now on, I assume that $\mathbb{N}$ is defined to be $\{1,2,3,\ldots\}$. I claim that there does not exist such a function $f$.
Suppose on the contrary that $f$ exists. Let $g:=\varphi\circ f$. Then, $g$ satisfies Cauchy's functional equation. That is, there exists $c\in\mathbb{N}$ such that $g(n)=cn$ for all $n\in\mathbb{N}$.
Let $d_1,d_2,\ldots,d_k$ be all natural numbers that divide $c$. Pick pairwise distinct primes $p_1,p_2,\ldots,p_{2k}$ that do not divide $c$. Consider the system of congruences
$$d_ix\equiv-1\pmod{p_{2i-1}p_{2i}}$$
for $i=1,2,\ldots,k$. This congruence has a unique solution
$$x\equiv x_0\pmod{P}\,,$$
where $P:=p_1p_2\cdots p_{2k}$ and $x_0\in\mathbb{Z}_{>0}$. By Dirichtlet's Theorem, there exists a prime natural number $p>c$ such that $p\equiv x_0\pmod{P}$, noting that $\gcd(x_0,P)=1$. It follows that the equation
$$\varphi(N)=cp$$
has no solution $N\in\mathbb{N}$. Thus, $\varphi\big(f(p)\big)=g(p)=cp$ is impossible.
Best Answer
Introduce a new function $g : \Bbb{R} \to \Bbb{R}$ by
$$g(x) = e^{-x} f(e^x).$$
Then satisfies the Cauchy's functional equation
$$g(x+y) = g(x) + g(y).$$
This equation is extensively studied, and even a mild regularity condition will force the solution to be of the form $g(x) = cx$. On the other hand, under the Axiom of Choice we can construct a solution which is not of this form.