These two identities are the "opposite ends" (in a sense I will explain below) of
$$ \frac1{n!} \int_0^t \frac{x^n}{e^x} \,dx
+ \frac1{e^t} \sum_{k=0}^n \frac{t^k}{k!}
= 1 \tag{1} $$
(Proof: By induction, integrating by parts. Just like the usual proof of your second identity, but with $\int_0^t$ instead of $\int_0^\infty$.) If $t\ge 0$, then both terms on the left of (1) are nonnegative, so (1) implies they are both in $[0,1]$. The first term on the left expresses the accuracy of the truncated integral as an approximation to $n!$; the second expresses the accuracy of the truncated Taylor series as an approximation to $e^t$. What (1) asserts is that these approximations are in a see-saw relationship: when one is good, the other is bad.
Your identities are the "opposite ends" of (1) in the following sense: for fixed $n$, as $t\to\infty$, the first term approaches $1$ and the second approaches $0$, yielding your second identity in the limit; for fixed $t$, as $n\to\infty$, the first term approaches $0$ and the second approaches $1$, yielding your first identity in the limit. That's a kind of duality, I suppose, though it leaves mysterious the apparent formal symmetry that prompted your question, where the factorial and exponential functions seem to trade places.
For other (sufficiently differentiable) functions, it is straightforward to generalize (1) to
$$ \int_0^t \frac{x^n f^{(n+1)}(-x)}{n!} \,dx
+ \sum_{k=0}^n \frac{t^k f^{(k)}(-t)}{k!}
= f(0) \tag{2} $$
but again, this loses the nice formal symmetry that prompted your question.
(That symmetry is suspiciously imperfect, though, in that both identities have $x^n$ in the numerator, even though the roles of $x$ and $n$ are reversed in the two identities.)