I will try to give a more general answer: given an operator $A$, how can we interpret a function of the operator $f(A)$?
There are two possible options:
i) if the function $f(x)$ admits to a power-series expansion about $x=0$, we can use this to define the $f(A) = \sum \frac{f^{(n)}(0)}{n!} A^n$ and we know how to take integer powers of operators - we just apply the operator $n$ times.
ii) if the operator $A$ can be diagonalized and has eigenstates $|a\rangle$ with eigenvalues $a$ where the eigenstates form a basis, then we can define $f(A)$ via its operation on the eigenstates $f(A)|a\rangle = f(a) |a\rangle$. Then, for any state $|\psi\rangle$ we can know how $f(A)$ acts upon it by expanding $|\psi\rangle$ in the eigenbasis $|\psi\rangle = \sum_a c_a |a\rangle$ and then acting with $f(A)$ on each of the states in the expansion.
If both $A$ has a basis of eigenvectors and $f(x)$ has a power series, it is easy to see that both definitions are consistent with one another. If none of the things apply, then the function may be ill-defined (for example - $\ln(a^{\dagger})$, where $a^{\dagger}$ is the Harmonic oscillator raising operator, is not well defined).
Now for the case at hand -- you have the function $U = \exp(-i H t/\hbar)$ of the Hamiltonian. As both definitions can be applied here, we can choose to focus on the second. It is clear the the eigenvectors of $U$ are the eigenvectors of $H$, and if $H$ has an eigenvalue $E$ when acting on a state $H|\psi_E\rangle = E |\psi_E\rangle$, then $U|\psi_E\rangle = \exp(-i H t/\hbar) |\psi_E\rangle = \exp(-iEt/\hbar)|\psi_E\rangle$, which gives you the answer.
Note that with these definitions the operator associated by Stone's theorem to $U$ is not exactly $H$ but apparently $-H.$
Remember $c(a\otimes b)=(ca)\otimes b=a\otimes (cb).$ You can distribute $-1/it$ into the sum and onto the correct factor. $$\lim_{t\to0}\sum_{j=1}^nU(t)\psi_1\otimes\cdots\otimes\frac{U_j(t)\psi_j-\psi_j}{-it}\otimes\cdots\otimes\psi_n$$
As $t\to0,$ $U(t)\to1,$ so in each term of the sum we have $U(t)\psi_1\otimes\cdots\otimes U(t)\psi_{j-1}\to\psi_1\otimes\cdots\otimes\psi_{j-1},$ and we also get $\frac{U_j(t)\psi_j-\psi_j}{-it}\to H\psi_j$ by definition of $H.$
A nicer way to view this is to note that the definition of $H$ is just $-iH=U'(0)$ where $'$ denotes the derivative (treating $U$ as a function from $\mathbb R$ to operators). Then the $n$-particle unitary $U_n(t)=U(t)\otimes\cdots\otimes U(t)$ (where the extension of $\otimes$ to operators is understood) produces the $n$-particle Hamiltonian $H_n(t)=(H\otimes1\otimes\cdots\otimes1)+(1\otimes H\otimes1\otimes\cdots\otimes1)+\cdots+(1\otimes\cdots\otimes1\otimes H)$ by the product rule (because $\otimes$ has the algebraic properties of a product).
Best Answer
Ths Schrodinger equation in general form reads $$i\hbar \dfrac{d}{dt}|\psi\rangle=H|\psi\rangle\tag{1}$$
where $H$ is the Hamiltonian of the system. This equation can be solved in terms of one time evolution operator which is exactly the one you ask about. The reason the parameter in it is time is because the operator in the exponent, $H$, is the Hamiltonian operator so that $|\psi(t)\rangle=U(t)|\psi_0\rangle$ solves (1).
So, asking why that parameter is time is tantamount to asking why the Schrodinger equation should govern the time evolution of a system. But then, a point to appreciate is that the Schrodinger equation is one of the postulates of Quantum Mechanics. As such it is one of the axioms defining the theory and not something we can prove from something else. It is what we postulate because we have derived predictions from it which were observed in experiment.
Nevertheless, one may still motivate the Schrodinger equation. In particular, one way is by taking inspiration from Classical Mechanics. The Schrodinger equation is exactly the statement that the Hamiltonian is the generator of time translations. But the same is true in Classical Mechanics. So one may postulate the quantum version by demading that, as in the classical theory, the Hamiltonian still generates time translation.