If we set $$\tilde{f}(x) := \lim_{h \downarrow 0} f(x+h),$$ then $\tilde{f}$ is (by assumption) well-defined. Moreover, straight-forward computations show that $\tilde{f}$ is right-continuous (see Lemma 3 below) and, hence, Borel measurable. If we can prove that the set
$$J:=\{x \in \mathbb{R}; \tilde{f}(x) \neq f(x)\} \tag{1}$$
is countable, then $g:=f-\tilde{f}$ is Borel measurable (see e.g. this proof). Consequently, $f=g+\tilde{f}$ is Borel measurable as sum of Borel measurable functions.
To prove that $J$ is countable we proceed as follows: For $x \in \mathbb{R}$ define the oscillations of $f$ at $x$ by
$$\omega(x) := \inf_{r>0} \omega_r(x) := \inf_{r>0} \left( \sup_{z \in B(x,r)} f(z) - \inf_{z \in B(x,r)} f(z) \right).$$
It is not difficult to see that $$\{\omega=0\} = \{x \in \mathbb{R}; \text{$x$ is a continuity point of $f$}\} \tag{2}$$
and therefore $J \subseteq \{\omega \neq 0\}$. Consequently, we are done if we can show that $\{\omega \neq 0\}$ is countable.
Lemma 1: For any $x \in \mathbb{R}$ and $n \in \mathbb{N}$ there exists $\delta(x)>0$ such that $\omega(y) \leq 1/n$ for all $y \in (x,x+\delta(x))$.
Proof: Fix $x \in \mathbb{R}$ and $n \in \mathbb{N}$. Since the limit $\tilde{f}(x) = \lim_{h \downarrow 0} f(x+h)$ exists, we have
$$\sup_{z \in (x,x+h)} f(z) \xrightarrow[]{h \to 0} \tilde{f}(x) \qquad \inf_{z \in (x,x+h)} f(z) \xrightarrow[]{h \to 0} \tilde{f}(x),$$
and so
$$\lim_{h \to 0} \left| \sup_{z \in (x,x+h)} f(z) - \inf_{z \in (x,x+h)} f(z) \right|=0;$$
in particular we can choose $\delta>0$ such that
$$\left| \sup_{z \in (x,x+\delta)} f(z) - \inf_{z \in (x,x+\delta)} f(z) \right| \leq \frac{1}{n}. \tag{3}$$
Now let $y \in (x,x+\delta)$. If we set $r := \min\{|y-x|,|y-(x+\delta)|\}$ then $B(y,r) \subseteq (x,x+\delta)$. In particular, we have
$$\sup_{z \in B(y,r)} f(z) \leq \sup_{z \in (x,x+\delta)} f(z) \qquad \inf_{z \in B(y,r)} f(z) \geq \inf_{z \in (x,x+\delta)} f(z) \tag{4}$$
which implies
$$\inf_{z \in (x,x+\delta)} f(z) \leq \inf_{z \in B(y,r)} f(z) \leq \sup_{z \in B(y,r)} f(z) \leq \sup_{z \in (x,x+\delta)} f(z).$$
On the other hand, we know from $(3)$ that
$$\sup_{z \in (x,x+\delta)} f(z) \leq \inf_{z \in (x,x+\delta)} f(z) + \frac{1}{n}.$$
Combining the two chains of inequalities we conclude that
$$\sup_{z \in B(y,r)} f(z) - \inf_{z \in B(y,r)} f(z) \leq \frac{1}{n},$$
i.e. $\omega_r(y) \leq 1/n$. In particular, $\omega(y) \leq 1/n$ which finishes the proof of the Lemma.
Lemma 2: $\{\omega \neq 0\}$ is countable.
Proof: Clearly, it suffices to show that $\{\omega > 1/n\}$ is countable for each $n \in \mathbb{N}$. For fixed $n \in \mathbb{N}$ denote by $\delta(x)$ the constant from the previous lemma. For each fixed $k \in \mathbb{N}$ and $N \in \mathbb{N}$ the set
$$B_{k,N} := \{x \in [-N,N] \cap \{\omega>1/n\}; \delta(x) \geq 1/k\}$$
is finite. Indeed: By the previous lemma, the distance between any two points in $B_{k,N}$ is at least $1/k$ and since the length of the interval $[-N,N]$ is $2N$, there can exist at most $2Nk+1$ points in $B_{k,N}$. This implies that
$$\{x \in \{\omega>1/n\}; \delta(x) \geq 1/k\} = \bigcup_{N \in \mathbb{N}} B_{k,N}$$
is countable which, in turn, implies that
$$\{\omega>1/n\} = \bigcup_{k \in \mathbb{N}} \{x \in \{\omega>1/n\}; \delta(x) \geq 1/k\}$$
is countable.
Edit: Following the comment to my answer, I add a proof for the right-continuity of $\tilde{f}$.
Lemma 3: $\tilde{f}$ is right-continuous.
Proof: Since $\tilde{f}(y) = \lim_{h \downarrow 0} f(y+h)$ we clearly have
$$\inf_{z \in (y,y+r)} f(z) \leq \tilde{f}(y) \leq \sup_{z \in (y,y+r)} f(z) \tag{5}$$
for any $y \in \mathbb{R}$ and $r>0$. For fixed $x \in \mathbb{R}$ and $\epsilon=1/n$ let $\delta=\delta(x)>0$ be as in (the proof of) Lemma 1. Using (5) for $y=x$ we find that
$$\inf_{z \in (x,x+\delta)} f(z) \leq \tilde{f}(x) \leq \sup_{z \in (x,x+\delta)} f(z).$$
On the other hand it follows from (4) and (5) that
$$\inf_{z \in (x,x+\delta)} f(z) \leq \inf_{z \in B(y,r)} f(z) \leq \tilde{f}(y) \leq \sup_{z \in B(y,r)} f(z) \leq \sup_{z \in (x,x+\delta)} f(z)$$
for any $y \in (x,x+\delta)$ where $r:=\min\{|y-x|,y-(x+\delta)|\}$. Combining both inequalities and using (3) we get
$$|\tilde{f}(x)-\tilde{f}(y)| \leq \left| \sup_{z \in (x,x+\delta)} f(z) - \inf_{z \in (x,x+\delta)} f(z) \right| \leq \frac{1}{n}$$
for all $y \in (x,x+\delta)$ which finishes the proof of the right-continuity of $\tilde{f}$.
Best Answer
Tha answer is no.
Every pointwise limit of continuous functions is of Baire class $1$ (this is just the definition), and this implies that the set of discontinuities of $f$ is not all of $\Bbb{R}$ by a Baire category argument (see this post Construction of a function which is not the pointwise limit of a sequence of continuous functions).
But for example the Borel function $\chi_{\Bbb{Q}}$ (indicator function of the rationals) is discontinuous everywhere, hence not a pointwise limit of continuous functions.