Addendum
There is a potential flaw in the original proof (below) which assumes that convergence to right-hand limits is uniform. Here is a different proof.
Any function of bounded variation has at most countably many jump discontinuities. There are two possibilities.
(1) There are finitely many jump discontinuities in some open neighborhood of $\lambda$. In this case, there is an interval $(\lambda - \delta, \lambda + \delta)$ where $f$ is continuous except possibly at $\lambda$. For $x \neq \lambda$ we have $f_R(x) = f(x+) = f_L(x) = f(x-) = f(x).$ By hypothesis $f_R$ is continuous at $\lambda$ and, therefore, $f_R(\lambda-) = f_R(\lambda+)$.
Thus $$f(\lambda-) = \lim_{x \to \lambda -}f(x) = \lim_{x \to \lambda -}f_R(x) = f_R(\lambda-) = f_R(\lambda+) = \lim_{x \to \lambda +}f_R(x) = \lim_{x \to \lambda +}f(x) = f(\lambda+),$$
and $f$ is continuous at $\lambda.$
(2) Jump discontinuities accumulate at $\lambda$. Let $(y_n)$ and $(z_n)$ be any increasing and decreasing sequences of points, respectively, both converging to $\lambda$. Since, the sums of the jumps must be bounded by the total variation of $f$, we have
$$\sum_{k=1}^\infty |f_R(y_k) - f_L(y_k)| < \infty, \\ \sum_{k=1}^\infty |f_R(z_k) - f_L(z_k)| < \infty,$$
and $\lim_{k \to \infty} |f_R(y_k) - f_L(y_k)| = \lim_{k \to \infty} |f_R(z_k) - f_L(z_k)| = 0$.
Thus,
$$f_L(\lambda-) = \lim_{k \to \infty} f_L(y_k) = \lim_{k \to \infty} f_R(y_k) - \lim_{k \to \infty} (f_R(y_k) - f_L(y_k)) = f_R(\lambda-), \\ f_L(\lambda+) = \lim_{k \to \infty} f_L(z_k) = \lim_{k \to \infty} f_R(z_k) - \lim_{k \to \infty} (f_R(z_k) - f_L(z_k)) = f_R(\lambda+), $$
and
$$f_L(\lambda-) = f_L(\lambda+) = f_R(\lambda+) = f_R(\lambda-).$$
Original Proof
By hypothesis, $\tilde{f}$ is continuous at $\lambda$. We must have $f(\lambda-) = f(\lambda+) =\tilde{f}(\lambda)$ when $ f$ is continuous at $\lambda$.
Suppose $f$ is not continuous at $\lambda$. This implies $f(\lambda-) \neq f(\lambda+) = \tilde{f} (\lambda).$ Since the left-hand limit exists there exists a sequence $(x_n)$ converging to $\lambda$ from the left such that $f(x_n)$ does not converge to $\tilde{f}(\lambda)$.
Hence, there exists $\epsilon_0 > 0$ such that for all sufficiently large $n$,
$$|f(x_n) - \tilde{f}(\lambda)| > \epsilon_0.$$
Now choose a sequence of points $(y_n)$ such that $x_n - 1/2^{n} < y_n < x_n.$ Then $y_n \to \lambda$ and $|f(x_n) - \tilde{f}(y_n)| < \epsilon_0/2$ for sufficiently large $n$, since the right-hand limit exists everywhere (and convergence is uniform in a compact neighborhood of $\lambda$).
Thus,
$$|\tilde{f}(y_n) - \tilde{f}(\lambda)| > |f(x_n) - \tilde{f}(\lambda)| - |f(x_n) - \tilde{f}(y_n)| > \epsilon_0/2, $$
which contradicts the continuity of $\tilde{f}$ at $\lambda.$
Best Answer
If $f$ is Riemann integrable on $[a,b]$, then there exists a partition $\mathcal{P}=\{a=x_0<\ldots<b=x_n\}$ such that $$U(f,P)-L(f,P)<\varepsilon$$ where $$\begin{align} U(f,P)&=\sum^n_{j=1}M_j(x_j-x_{j-1})\\ L(f,P)&=\sum^n_{j=1}m_j(x_j-x_{j-1}) \end{align}$$ where $M_j=\sup_{x\in[x_{j-1},x_j]}f(x)$, and $m_j=\inf_{x\in[x_{j-1},x_j]}f(x)$. Define the step functions $$\begin{align} g(x)&:=\sum^n_{j=1}m_j\mathbb{1}_{(x_{j-1},x_j]}(x)\\ h(x)&=\sum^n_{j=1}M_j\mathbb{1}_{(x_{j-1},x_j]}(x) \end{align}$$
Clearly $g\leq f\leq h$, and $L(f,P)=\int^b_a g\leq \int^b_a f\leq \int^b_a h= U(f,P)$. The rest follows easily: $$\begin{align} \int^b_a (f -g)&\leq U(f,P)-L(f,P)<\varepsilon\\ \int^b_a (h -f)&\leq U(f,P)-L(f,P)<\varepsilon \end{align}$$