It's impossible to say with confidence that the two distributions are exactly the same, without making some fairly strong restrictions about what the two distributions could look like. Imagine a case where you want to compare a distribution $P$ to a distribution $Q = (1 - \varepsilon) P + \varepsilon Z$, where $\varepsilon$ is say $10^{-6}$. You probably won't even have any samples from $Z$ until you take about $10^{6}$ samples; even then, if the handful of samples you got from $Z$ are even vaguely plausible under $P$, it'll look fine.
What you can do is test this: "$P$ and $Q$ are almost the same." This sounds like what you want ("I am trying to show that under those settings, for all practical purposes we can treat the distributions as the same"). The problem is, what does "almost the same" mean? You'll need to specify some kind of distance between $P$ and $Q$.
The thing that makes this somewhat more complicated than the usual KS test-style setup is that we often can only easily derive the (asymptotic) distribution of whatever handy test statistic under the case when $P = Q$, but if we want our null distribution to be $P \ne Q$, then that doesn't help.
One natural distance is the total variation distance, which can be defined as
$$
\operatorname{TV}(P, Q)
= \sup_{f : \sup_x \lvert f(x) \rvert \le 1}
\operatorname{\mathbb E}_{X \sim P}[ f(X) ]
- \operatorname{\mathbb E}_{Y \sim Q}[ f(Y) ]
= \sup_{\mathcal E} \lvert P(\mathcal E) - Q(\mathcal E) \rvert
.$$
The total variation is the most that the two distributions can disagree on the probability of two events;
equivalently, if you apply a function $f$ to the two distributions that is never large,
it's the most that function $f$ can differ between the two distributions.
For discrete distributions, the following paper did something like what you want:
Chan, Diakonikolas, G. Valiant, and P. Valiant. Optimal Algorithms for Testing Closeness of Discrete Distributions. SODA 2013. (arXiv)
This paper gives a rate-optimal algorithm which distinguishes the case $P = Q$ from the case $\operatorname{TV}(P, Q) \ge \varepsilon$. That is, if you give it samples from two $P$ and $Q$ supported on an $\ell$-element discrete set which are at least $\varepsilon$ apart in total variation, then it'll correctly tell you that they're apart with high probability if you give it at least $\Omega(\max\{ n^{2/3} \varepsilon^{-4/3}, \ell^{1/2} \varepsilon^{-2} \})$ samples; likewise, if you give it two samples that are the same, it'll tell you with high probability that they're the same with enough samples. But if you give it distributions which are $\varepsilon/2$ apart in total variation, it might give you either answer (until you give it enough samples to distinguish with $\varepsilon/2$).
Unfortunately, as this is a theory paper, one of the inputs to the algorithm is "a suitable constant $C$." I'm not super familiar with this subarea, but I don't know if there is an implementable version of this algorithm that gives you something usable as a classical statistical test.
Another nice distance is the maximum mean discrepancy (MMD). This distance can be defined as
$$
\operatorname{MMD}(P, Q)
= \sup_{f : \lVert f \rVert_{\mathcal H} \le 1}
\operatorname{\mathbb E}_{X \sim P}[ f(X) ]
- \operatorname{\mathbb E}_{Y \sim Q}[ f(Y) ]
,$$
where $\lVert f \rVert_{\mathcal H}$ is the norm in a reproducing kernel Hilbert space. This means that if the MMD is small, applying any "smooth" function to the two distributions (where "smooth" is defined in terms of the RKHS kernel) will get you similar answers. MMD testing is usually done with the classic null hypothesis of $P = Q$; the standard reference is
Gretton, Borgwardt, Rasch, Schölkopf, and Smola. A Kernel Two-Sample Test. JMLR 13(Mar):723–773, 2012. (publisher page)
But, since I know more about MMD testing than I do about total variation on discrete distributions, let's figure out a test for the hypotheses
\begin{gather}
H_0 : \operatorname{MMD}(P, Q) > \varepsilon
\qquad
H_1 : \operatorname{MMD}(P, Q) < \varepsilon
\tag{1}
.\end{gather}
From Gretton et al. (2012), we know that the following is a simple estimator of the MMD: let $k$ be the kernel of the RKHS $\mathcal H$, and suppose we have $\{X_i\}_{i=1}^m \sim P^m$, $\{Y_i\}_{i=1}^n \sim Q^n$. Then the following is a reasonable estimator for $\operatorname{MMD}$:
$$
\operatorname{MMD}_b^2(X, Y)
= \frac{1}{m^2} \sum_{i=1}^m \sum_{j=1}^m k(X_i, X_j) + \frac{1}{n^2} \sum_{i=1}^m \sum_{j=1}^n k(Y_i, Y_j) - \frac{2}{mn} \sum_{i=1}^m \sum_{j=1}^n k(X_i, Y_j)
.$$
In particular, a one-sided analogue to their Theorem 7 tells us (via McDiarmid's inequality) that, when $0 \le k(x, y) \le K$ for all $x, y$,
$$
\Pr\left( \operatorname{MMD}(P, Q) - \operatorname{MMD}_b(X, Y) > 2 \left( \sqrt{\frac{K}{m}} + \sqrt{\frac{K}{n}} \right) + t \right)
\le \exp\left( \frac{- t^2 m n}{2 K (m + n)} \right)
\tag{2}
.$$
Thus we can say, with no assumptions about $P$ or $Q$ and holding exactly for all $m$ and $n$,
with probability at most $\delta$ the true value of $\operatorname{MMD}(P, Q)$ is at most
$$
\operatorname{MMD}_b(X, Y)
+ 2 \sqrt{K} \left( \frac{1}{\sqrt{m}} + \frac{1}{\sqrt{n}} \right)
+ \sqrt{2 K \left( \frac{1}{m} + \frac{1}{n} \right) \log\frac1\delta}
;$$
this tells you how "almost-the-same" you can confidently say the two distributions are.
If you'd prefer the hypothesis testing setup (1), you can easily manipulate (2) into giving you that as well: your $p$-value is just
$$
\exp\left( \frac{- \left( \max\left\{ 0, \varepsilon - \operatorname{MMD}_b(X, Y)
- 2 \sqrt{K} \left( \frac{1}{\sqrt{m}} + \frac{1}{\sqrt{n}} \right) \right\} \right)^2}{2 K \left( \frac1m + \frac1n \right)} \right)
.$$
You could also probably use the asymptotic normal distribution of the MMD estimator to get a less-conservative asympotically-valid test. But I just noticed that this question is two years old, and so I won't write that out unless someone asks. :)
