On leo's request I'm posting my comment as an answer.
Your treatment of the equality cases of Hölder's and Minkowski's inequalities are perfectly fine and clean. There's a small typo when you write that $\int|fg| = \|f\|_p\|g\|_q$ if and only if $|f|^p$ is a constant times of $|g|^q$ almost everywhere (you write the $p$-norm of $f$ and the $q$-norm of $g$ instead).
The case where either one $\|f\|_p$ or $\|g\|_q$ (or both) are infinite isn't part of this exercise and simply wrong. You can trisect $E = F \cup G \cup H$ into disjoint measurable sets of positive measure, take $f$ not $p$-integrable on $F$ and zero on $G$, take $g$ not $q$-integrable on $G$ and zero on $F$ and choose $fg$ non-integrable on $H$. Then certainly no power of $|f|$ is a constant multiple of a power of $|g|$ and vice versa, even though equality holds in the Hölder inequality.
A very nice “blackboard summary” of the equality case (for finite sequences) is given in Steele's excellent book The Cauchy–Schwarz Master Class. Let $a = (a_1,\ldots,a_n) \geq 0$ and $b = (b_1, \ldots, b_n) \geq 0$ and let $\hat{a}_i = \dfrac{a_i}{\|a\|_p}$ and $\hat{b}_i = \dfrac{b_i}{\|b\|_q}$. Then your argument is subsumed by the diagram (with an unfortunate typo in the upper right corner—no $p$th and $q$th roots there):
Mimicking this for functions, let us write $\hat{f} = \dfrac{|f|}{\|f\|_p}$ and $\hat{g} = \dfrac{|g|}{\|g\|_q}$ (assuming of course $\|f\|_p \neq 0 \neq \|g\|_q$), so $\int \hat{f}\vphantom{f}^p = 1$ and $\int \hat{g}^q =1$ and thus your argument becomes
$$
\begin{array}{ccc}
\int |fg| = \left(\int|f|^p\right)^{1/p} \left(\int|g|^q\right)^{1/q} & & |f|^p = |g|^q \frac{\|f\|_{p}^p}{\|g\|_{q}^q} \text{ a.e.}\\
\Updownarrow\vphantom{\int_{a}^b} & & \Updownarrow \\
\int \hat{f}\,\hat{g} = 1 & & \hat{f}\vphantom{f}^p = \hat{g}^q \text{ a.e.} \\
\Updownarrow\vphantom{\int_{a}^b} & & \Updownarrow \\
\int \hat{f}\,\hat{g} = \frac{1}{p} \int \hat{f}\vphantom{f}^p + \frac{1}{q} \int \hat{g}^q & \qquad \iff \qquad &
\hat{f}\,\hat{g} = \frac{1}{p} \hat{f}\vphantom{f}^p + \frac{1}{q} \hat{g}^q \text{ a.e.}
\end{array}
$$
I suggest that you draw a similar diagram for the equality case of Minkowski's inequality.
Let me write out my comment so that this question gets an answer. The dominated convergence theorem says the following.
If a sequence of measurable functions $\displaystyle \{f_n(x)\}_{n=1}^{\infty}$, on the measure space $(\Omega, \mathcal{F}, \mu)$, converge point-wise to a function $f(x)$ and every $f_n(x)$ is dominated by some integrable function $g(x)$ i.e. $\lvert f_n(x) \rvert \leq g(x)$, $\forall n$, $\forall x \in \Omega$ and $\displaystyle \int_{\Omega} g d\mu < \infty $, then $\displaystyle \int_{\Omega} f d \mu$ exists and $$\lim_{n \rightarrow \infty} \int_{\Omega} f_n d \mu = \int_{\Omega} f d \mu$$
Note that it is important that the dominating function $g$ is integrable as Henry T. Horton points out in the comments. Also, the condition $\lvert f_n \rvert \leq g$ everywhere, can be relaxed to $\lvert f_n \rvert \leq g$ $\mu$-almost everywhere.
Also, we need $s > -1$. Otherwise the integrals given in the question diverge.
Now lets apply this to the problem at hand. We want to evaluate $\displaystyle \lim_{n \rightarrow \infty} \int_0^{n} \left( 1 - \frac{x}n\right)^n x^s dx$.
Define $$f_n(x) = \begin{cases} \left( 1 - \frac{x}n\right)^n x^s & \text{ if }x \in [0, n]\\ 0 & \text{ otherwise}\end{cases}$$
Hence, $\displaystyle \lim_{n \rightarrow \infty} \int_0^{n} \left( 1 - \frac{x}n\right)^n x^s dx = \lim_{n \rightarrow \infty} \int_0^{\infty} f_n(x) dx$.
Now note that $f_n(x)$ converges point-wise to $e^{-x} x^s$ on $[0, \infty)$. This is so since $$\displaystyle \lim_{n \rightarrow \infty} \left(1 - \frac{x}{n} \right)^n = \exp(-x).$$
More importantly, $f_n(x)$ is dominated by $e^{-x} x^s$ on $[0,\infty)$ i.e. $\lvert f_n(x) \rvert \leq e^{-x} x^s$. This follows from the fact that $$1 - t \leq \exp(-t)$$ whenever $0 \leq t \leq 1$. Hence, we get that $$\left(1 - \frac{x}{n} \right) \leq \exp \left(-\frac{x}{n} \right)$$ which in-turn gives us $$\left(1 - \frac{x}{n} \right)^n \leq \exp \left(-x \right).$$
Hence, $\displaystyle f_n(x) \leq \exp(-x) x^s $. Also, $\displaystyle \int_0^{\infty} \exp(-x) x^s dx < \infty$ for all $s > -1$. Hence, in this case, the dominating function is the same as the limit function.
Putting all these things together, we get the desired result.
\begin{align}
\displaystyle \lim_{n \rightarrow \infty} \int_0^{n} \left( 1 - \frac{x}n\right)^n x^s dx & = \lim_{n \rightarrow \infty} \int_0^{\infty} f_n(x) dx & \text{(From the definition of $f_n(x)$)}\\
& = \int_0^{\infty} \lim_{n \rightarrow \infty} f_n(x) dx & \text{(Since $f_n(x)$ is dominated by $f(x)$)}\\
& = \int_0^{\infty} f(x) dx & \text{(Since $\displaystyle \lim_{n \rightarrow \infty} f_n(x) = f(x)$)}\\
& = \int_0^{\infty} \exp(-x) x^s dx
\end{align}
Best Answer
Consider using Fatou's lemma instead. Let $s_n := \left|\sum\limits_{k = 1}^n f_k\right|^p$. By Fatou's lemma, $$\left\|\sum_{n = 1}^\infty f_n\right\|_p^p = \|\lim_{n\to \infty} s_n\|_1 \le \varliminf_{n\to \infty} \|s_n\|_1 = \varliminf_{n\to \infty} \left\|\sum_{k = 1}^n f_k\right\|_p^p.$$ Therefore $$\left\|\sum_{n = 1}^\infty f_n\right\|_p \le \varliminf_{n\to \infty} \left\|\sum_{k = 1}^n f_k\right\|_p.$$ By Minkowski's inequlity,
$$\varliminf_{n\to \infty} \left\|\sum_{k = 1}^n f_k\right\|_p \le \varliminf_{n\to \infty} \sum_{k = 1}^n \|f_k\|_p = \sum_{n = 1}^\infty \|f_n\|_p.$$
Hence
$$\left\|\sum_{n = 1}^\infty f_n\right\|_p \le \sum_{n = 1}^\infty \|f_n\|_p.$$