As I commented under you post, I will use "$f(x-)$" instead of "$f(x+)$" for the left limit of a function $f$ in the proof below.
Preparations
The key of the proof is a property of the quantile function of a random variable $X$. In general, if $X$ is a random variable with distribution function $F$, we can define its quantile function (see, e.g., Equation (14.5) in Probability and Measure by Patrick Billingsley) of $X$ as
\begin{align}
\varphi(u) = \inf[x: F(x) \geq u], \quad 0 < u < 1.
\end{align}
For fixed $u \in (0, 1)$, we will need the following property of $\varphi$ (the proof of it can also be found in the reference mentioned above):
\begin{align}
& x \geq \varphi(u) \iff F(x) \geq u, \tag{1} \\
& x < \varphi(u) \iff F(x) < u. \tag{2}
\end{align}
As a rigorous proof, some properties of measure-theoretic integral will also be used tacitly. If you have questions, please comment below the answer.
Proof
Denote $F^\star(X; V) = F(X-) + V(F(X) - F(X-))$ by $U$, to show $U \sim U(0, 1)$, it suffices to show for any $u \in (0, 1)$, it holds that $P[U < u] = u$.
To this end, let's denote the set of discontinuities of $F$ by $J$, it then follows by the independence between $X$ and $V$ and $(2)$ that
\begin{align}
& P[U < u] = P[U < u, X \in J] + P[U < u, X \in J^c] \\
=& P[F(X-) + V(F(X) - F(X-)) < u, X \in J] + P[F(X) < u, X \in J^c] \\
=& \int_J P\left[V < \frac{u - F(x-)}{F(x) - F(x-)}\right]dF(x) +
P[X < \varphi(u), X \in J^c] \\
=& \int_J P\left[V < \frac{u - F(x-)}{F(x) - F(x-)}\right]dF(x) +
\int_{J^c \cap (-\infty, \varphi(u))} dF(x). \tag{3}
\end{align}
Let's analyze the integrand of the first term in the right-hand of $(3)$: If $x \in (-\infty, \varphi(u)) \cap J$, then $F(x-) < F(x) < u$ by the monotonicity of $F$ and $(2)$, whence $\frac{u - F(x-)}{F(x) - F(x-)} > 1$ and it then follows by $V \sim U(0, 1)$ that $P\left[V < \frac{u - F(x-)}{F(x) - F(x-)}\right] = 1$; if $x \in (\varphi(u), +\infty) \cap J$, then for every $t \in (\varphi(u), x)$, by $(1)$ we have $F(t) \geq F(\varphi(u)) \geq u$, hence $F(x-) = \lim_{t \to x^-}F(t) \geq u$, whence $\frac{u - F(x-)}{F(x) - F(x-)} \leq 0$ and it then follows by $V \sim U(0, 1)$ that $P\left[V < \frac{u - F(x-)}{F(x) - F(x-)}\right] = 0$. This implies that (decompose $J$ into $J = [J \cap (-\infty, \varphi(u))] \cup [J \cap \{\varphi(u)\}] \cup [J \cap (\varphi(u), +\infty)]$):
\begin{align}
& \int_J P\left[V < \frac{u - F(x-)}{F(x) - F(x-)}\right]dF(x) \\
=& \int_{J \cap (-\infty, \varphi(u))} dF(x) + \int_{J \cap \{\varphi(u)\}}
P\left[V < \frac{u - F(x-)}{F(x) - F(x-)}\right]dF(x). \tag{4}
\end{align}
$(3)$ and $(4)$ together then give
\begin{align}
P[U < u] = \int_{(-\infty, \varphi(u))} dF(x) + \int_{J \cap \{\varphi(u)\}}
P\left[V < \frac{u - F(x-)}{F(x) - F(x-)}\right]dF(x). \tag{5}
\end{align}
To simplify $(5)$, consider the following two cases:
Case 1: $\varphi(u) \in J^c$, i.e., $F$ is continuous at $\varphi(u)$. In this case the second term in the right-hand side of $(5)$ vanishes, and $F(\varphi(u)) = F(\varphi(u)-)$. $(5)$ thus reduces to
\begin{align}
P[U < u] = \int_{(-\infty, \varphi(u))} dF(x) = P[X < \varphi(u)] =
F(\varphi(u)-) = F(\varphi(u)) = u.
\end{align}
Case 2: $\varphi(u) \in J$, i.e., $F$ jumps at $\varphi(u)$. In this case the second term in the right-hand side of $(5)$ becomes
\begin{align}
& \int_{\{\varphi(u)\}}P\left[V < \frac{u - F(x-)}{F(x) - F(x-)}\right]dF(x) \\
=& P\left[V < \frac{u - F(\varphi(u)-)}{F(\varphi(u)) -F(\varphi(u)-)}\right] \times (F(\varphi(u)) -F(\varphi(u)-)) \\
=& u - F(\varphi(u)-).
\end{align}
Therefore $(5)$ reduces to
\begin{align}
P[U < u] = P[X < \varphi(u)] + u - F(\varphi(u)-) =
F(\varphi(u)-) + u - F(\varphi(u)-) = u.
\end{align}
This completes the proof.
