For consistency of terminology, let me say that $X_t$ and $Y_t$ are $M$-equivalent if one is a modification of the other, and $D$-equivalent if they are indistinguishable from one another. (This is not standard terminology, but I find the difference in syntax between the two terms makes it slightly difficult to write.)
Here's a way of looking at it based on sampling.
Suppose $X_t,Y_t$ are $M$-equivalent. Now choose $t_1 \in I$ arbitrarily. Repeatedly run $X_t$ and $Y_t$ "independently", but only sample them at time $t_1$. Then as you sample more and more times, the fraction of the samples such that $X_{t_1}=Y_{t_1}$ will converge to $1$, regardless of which $t_1$ you chose.
Suppose now that they are $D$-equivalent. Repeatedly run $X_t$ and $Y_t$ again, but this time record the entire trajectory. Then as you sample more and more times, the fraction of the samples such that $X_t$ and $Y_t$ are equal at every time will converge to $1$.
So $D$-equivalence automatically implies $M$-equivalence, since almost all samples have equality at every time and hence at any particular time.
When $I$ is countable, $M$-equivalence also implies $D$-equivalence, because the event "$X_t=Y_t$ for every $t \in I$" is the intersection of the events "$X_{t_k}=Y_{t_k}$" over $t_k \in I$, and each of these has probability $1$.
But when $I$ is uncountable (as in problems in continuous time), we may have $M$-equivalence but not $D$-equivalence. To see this, suppose $X_t,Y_t$ are $M$-equivalent and let $N(t)$ be the event that $X_t \neq Y_t$. Then $N(t)$ has probability zero. (If you like, this is just saying that $\int_t^t dx = 0$.) Then we are in $N=\bigcup_{t \in I} N(t)$ if there is some $t$ such that $X_t \neq Y_t$. Now $N$ is an uncountable union of sets of probability zero. So it might have positive probability, or it might not even be measurable.
Let's imagine sampling from the example from Oksendal. Pick a $t_1 \in [0,1]$, now the random time will be $t_1$ only with probability zero. So the fraction of samples with $X_{t_1} \neq Y_{t_1}$ will get smaller as we take more and more samples. But if we look at the whole trajectory instead, then at some random time we will always have $X_t \neq Y_t$, in every single sample. We will never see the exact same trajectory from both.
It is true that the paths of $Y$ all have only one point of discontinuity (namely the value of $T$), but that still means that all paths of $Y$ are not (right-)continuous, so in fact $Y$ is right-continuous with probability $0$, which is why the result you cited does not apply here.
You were confusing null sets $B\subset \mathbb{R}$ with respect to Lebesgue measure and null sets $N\subset \Omega$ with respect to the probability measure $\mathbb{P}$ (where $\mathbb{P}$ is defined on some measurable space $(\Omega,\mathcal{A})$).
Best Answer
Let $A=\bigcap_i \{X(t_i)=Y(t_i)\}$ Then $P(A)=1$ and $(X_{t_1},X_{t_2},\dots,X_{t_n})=(Y_{t_1},Y_{t_2},\dots,Y_{t_n})$ on $A$. Hence $P((X_{t_1},X_{t_2},\dots,X_{t_n}) \in C)=P((Y_{t_1},Y_{t_2},\dots,Y_{t_n}) \in C)$ for any Borel set $C$ in $ E^{n}$.