$\newcommand{\ve}{\varepsilon}$ I'll try to give an $\ve$-$\delta$ proof.
You assume that $f(x+)$ exists for each $x$. You have defined $g[a,b]\to\mathbb{R}:x\mapsto\begin{cases}f(x+)&a\le x\lt b\\ f(b)&x=b\end{cases}$ and you're asking whether at each point both one-sided limits $g(x-)$ and $g(x+)$ exist and whether $g(x+)=g(x)$.
Clearly, if we show that $f(x+)=g(x+)$ and $f(x-)=g(x-)$, then this is true.
Fix a point $x_0\in[a,b)$
Let $r=f(x_0+)$, i.e. for each $\ve>0$ there is a $\delta>0$ such that $|f(x)-r|<\ve$ for each $x\in(x_0,x_0+\delta)$. This clearly implies that $|g(x)-r|=|f(x+)-r|\le\ve$ for each $x\in(x_0,x_0+\delta/2)$. This shows that $g(x_0+)=r=f(x_0+)$.
Let $l=f(x_0-)$, i.e. for each $\ve>0$ there is a $\delta>0$ such that $|f(x)-l|<\ve$ for each $x\in(x_0-\delta,x_0)$. Now if $x\in(x_0-\delta,x_0)$ then we have an interval $(x,x_0)$ on the right from $x$ such that $|f(x')-l|<\ve$ for each $x'$ in this interval. From this we get $|g(x)-l|=|f(x+)-l|\le\ve$ for each $x\in(x_0-\delta,x_0)$. This shows that $g(x_0-)=l=f(x_0-)$
There is a result which says that for any real function there exist only countably many points $x$ of $\mathbb R$ for which $f$ is not continuous at $x$ but $f(x+)$ exists. See e.g. van Rooij, Schikhof: A Second Course on Real Functions, Theorem 7.7, p.45.
Clearly $f(x)\ne g(x)=f(x+)$ implies that $f$ is not continuous at $x$. By the above result, there is only countably many such points.