I suspect the confusion may be in what you mean by "step" in the phrase "$N$ steps ahead".
If you mean the one-second time step of your discrete time Markov process, then what you're calculating is exactly correct — with a $0.98$ self-loop transition probability, the process has a less than $0.5$ probability of moving anywhere within $N < 35$ steps.
If, however, by "step" you mean a transition in which the process does not stay in the same state, then you need to eliminate the self-loop transitions from your matrix (since you don't count them as steps) by setting their probability to $0$ and normalizing the remaining transition probabilities for each state so that they again sum to $1$.
Of course, what you lose by doing so is the information about the residence time in each state. If you wanted to know where the process will be after $N$ non-self-loop steps and how long it'll take to get there, that would require a rather more complicated calculation, and the resulting discrete phase-type distribution would, in general, be unlikely to have a simple form (although it might be possible to approximate it with some nicer distributions).
In the comments, you write that your transition matrix is (approximately) $$P=\pmatrix{0.99&0.0208&0&0\\0.0019&0.9792&0.0104&0\\0&0&0.9896&0.0108\\0.0028&0&0&0.9907}$$ and that "second option it is what I'm looking for", i.e. that you want to calculate the probability distribution after $N$ steps, a step denoting a transition between two distinct states. If so, what you need to do is to calculate another stochastic matrix $P'$ which describes the same process, but which does not include the self-loop transitions (i.e. whose diagonal entries are zero).
To obtain $P'$ from $P$, simply zero out the diagonal entries and re-normalize each row so that the matrix stays stochastic (i.e. so that each row sums to $1$). Since most of the rows in $P$ only have one non-zero off-diagonal element to begin with, that element becomes $1$, leaving row 2 as the only non-trivial case: $$P' = \pmatrix{0&1&0&0\\0.1545&0&0.8455&0\\0&0&0&1\\1&0&0&0}$$ From this, you can calculate the state distribution $\pi_n = \pi P'^n$ after $n$ steps, given a starting distribution $\pi$.
However, note that, whereas the original transition matrix $P$ was ergodic, $P'$ isn't — in particular, a process in an even-numbered state always ends up in an odd-numbered state after one step and vice versa, so all states have an even period. This lack of ergodicity doesn't really affect the calculation of $\pi_n$ as such, but it does mean e.g. that $\pi_n$ won't generally converge to a stationary distribution as $n \to \infty$.
Let $(X_t)_{t\geq 0}$ be the process given by the second construction, driven by a Poisson process $(N_t)_{t\geq 0}$ of rate $\lambda$ and a DTMC $(Y_n)_{n\in\mathbb{N}}$ with transition matrix $Q$.
Fix $h>0$. Let's start $X$ in state $i$ (i.e. $X_0=Y_0=i$) and consider the probability of it being at state $j$ (with $j\neq i$) at time $h$.
$$\begin{align}\mathbb{P}[X_h=j] &= \sum_{n\in\mathbb{N}} \mathbb{P}[X_h=j, N_h=n]\\
&= \mathbb{P}[X_h=j, N_h=1] + o(h)\\
&= \mathbb{P}[Y_1=j]\mathbb{P}[N_h=1] + o(h)\\
&= \frac{g_{ij}}{\lambda}(\lambda h)+ o(h)\\
&= g_{ij}h+o(h)
\end{align}
$$
This is one characterization of a CTMC with generator $G$.
Intuitive remarks: Since $\lambda$ cancels, its precise value doesn't matter. In the second construction more "jumps" occur, because of the condition on $\lambda$, but the DTMC is able to transition to its current state, and these jumps aren't observed in the resultant CTMC.
Best Answer
Well, let's try this... Let $\{N_{t}\}_{t\geq0}$ be an inhomogeneous Poisson process with rate function $\{\lambda(t)\}_{t\geq0}$. Let's try to compute the transition probabilites.
By definition of an inhomogeneous Poisson process, we have
$$P\{N(t+h)-N(t)=1\}=\lambda(t)h+o(h)$$ $$P\{N(t+h)-N(t)>1\}=o(h)$$
Since $\{N_{t}\}_{t\geq0}$ has independent increments by definition, we have
$$P\{N(t+h)-N(t)=1|N(t)=k\}=\lambda(t)h+o(h)$$ $$P\{N(t+h)-N(t)>1|N(t)=k\}=o(h)$$
for any $t\geq0$, any $k\in\mathbb{N}_{0}$ and "small" $h$.
Dividing previous equations by $h$, and letting $h\rightarrow 0$, we get
$$\lim_{h\rightarrow0}\frac{1}{h}P\{N(t+h)-N(t)=i+1|N(t)=i\}=\lambda(t)$$ $$\lim_{h\rightarrow0}\frac{1}{h}P\{N(t+h)-N(t)=j|N(t)=i\}=0$$ for $j>i+1$ or $j<i$.