Your argument has a technical gap in that if $k=1$, then the existence of a subgroup of order $p^{k-1}$ of $G/B$ does not follow by the induction hypothesis, which is stated only for groups of order $p^k$ with $k\geq 1$. Of course, it is trivial to establish the desired resuld when $k=1$, but you did not do so correctly. In doing inductive arguments, it is very important to make sure the argument holds for all values of the variables, lest you end up proving that all horses are of the same color. There is also a problem that you seem to phrase your result with both $p$ and $k$ fixed, in which case you cannot apply the Induction Hypothesis to $G/B$, because you may not have $p^k\mid |G/B|$. Instead, you seem to be proving a more general statement about prime powers dividing the order of $G$, rather than the specific prime power $p^k$ dividing the order of $G$.
So let's be clear about what we are trying to prove:
Let $p$ be a prime, and let $G$ be a finite group. If $k\geq0$ and $p^k\mid |G|$, then $G$ has a subgroup of order $p^k$.
You want to prove it by induction on $|G|$. This should be done by so-called "strong induction": assume the result holds for all groups of order strictly less than $n$, then prove it for groups of order $n$.
The reason, I think, that you are having trouble pin-pointing a proper "base case" is that in strong induction you do not need a base case! (see my previous discussions about this here and here). Instead, we just need to make sure that our general inductive argument is valid for all possible values of $n$; if it breaks down for some, we will need to establish the theorem for those values as "special cases", not as a basis for induction.
So, let us clean up your argument a bit with this in mind. We will assume the following induction hypothesis:
Induction Hypothesis. For any group $K$ of order strictly smaller than $n\geq 1$, if some power of $p$ divides $|K|$, then $K$ has a subgroup of that order.
We now let $G$ be a group of order $n$, and we wish to prove that if $p^k\mid |G|$ for some $k\geq 0$, then $G$ has a subgroup of order $p^k$. Because we are trying to prove an implication, we may assume that $p^k\mid |G|=n$.
If $k=0$, then the trivial group shows the existence of a subgroup of order $p^k$. (This is how we bypass the issue you had about trying to apply an induction hypothesis that only works for $k\geq 1$ to a group that may not be divisible by $p$.)
So we may assume $k\gt 0$.
We do not need to consider the case of $G$ abelian and nonabelian separately. As you do, let us look at the class equation for $G$: there exist noncentral $w_1,\ldots,w_m$, $m\geq 0$ (I'm changing to $m$ because I'm using $n$ to denote $|G|$) such that
$$|G| = |Z(G)| + \sum_{i=1}^m [G:C_G(w_i)].$$
It's possible that $m=0$, in which case $G$ is abelian.
You argue: if there exist an $i$, $1\leq i\leq m$, such that $p^k\mid |C_G(w_i)|$, then because $w_i$ is not central we know $C_G(w_i)\neq G$, so the order of $C_G(w_i)$ is strictly smaller than $n$. Applying the induction hypothesis to $C_G(w_i)$ we obtain a subgroup $P\leq C_G(w_i) \leq G$ of order $p^k$, and we are done. This is correct, and a correct invocation of the induction hypothesis. (If $m=0$, then of course this cannot happen and we will fall into the alternative case.)
If this is not the case, then $p$ must divide $[G:C_G(w_i)]$ for each $i$, $1\leq i\leq m$, and therefore $p$ must divide $|Z(G)|$, since $p$ also divides $|G|$. (This is also true if $m=0$, by the way; that's why we don't need to argue about abelian vs. nonabelian $G$). Then $Z(G)$ has a subgroup of order $p$ (by Cauchy's Theorem), say $B$. This is central in $G$, hence normal. So consider now $G/B$. This is a group of order strictly smaller than $n$ (namely, $\frac{n}{p}$), and therefore the induction hypothesis applies: if some power of $p$ divides $|G/B|$, then $|G/B|$ has a subgroup of that order. Since $k\gt 0$, we can talk about $p^{k-1}$ as a "power of $p$"; we now note that $p^{k-1}$ divides $|G/B|$, then $G/B$ has a subgroup $A$ of order $p^{k-1}$. This $A$ corresponds to a subgroup $H$ of $G$ that contains $B$, $A=H/B$, and thus the order of $H$ is $p^k$, proving that $G$ has a subgroup of order $p^k$. This is also a correct application of the induction hypothesis and of the correspondence theorem.
And we are done!
Okay, now let's go over the argument and make sure it works for any value of $n$. It does, so we do not need to do a "special case" (often mis-identified as a "base case"). So, by strong induction, the result holds. $\Box$
Now, if there is an issue with the proof I would say that I would be a bit uneasy at first in that we are really making a kind of "double induction" argument, with one induction on $k$ (having established the case of $p^1$ via Cauchy's Theorem), and another induction on $|G|$. Which is somewhat obscured in how you phrased the result: you seem to fix $k$, in which case you cannot apply the induction hypothesis to $G/B$, because the induction hypothesis is for groups whose order is divisible by $p^{k}$, not for groups whose order is divisible by $p^{k-1}$; that's why it is important to state the result very explicitly so one is clear about what the induction hypothesis is and what one needs to establish at $n$ to ensure the inductive step is complete.
