I think $\exp(t)$ converges for every possible input of $t$, but I haven’t found a proof yet. If anyone can provide a proof or disproof then that would be helpful.
The proof may depend on the type of the input. If $t$ is a real/complex number, then convergence criteria like the ratio test may be applied (or, equivalently, calculating the radius of convergence). If $t$ is an element of some complete normed $\mathbb{R}$- (or $\mathbb{C}$-) vector space, then it suffices to show that the series is absolutely convergent (see e.g. here).
I would also like to know if for some $\exp(t)$ where $t \in U$, $\exp(t)$ will also be $\in U$.
This depends on what $U$ is. For example, if $U = [1, 2]$, this is not the case, since $2 \in U$ but $\exp(2) \notin U$. Another counterexample would be $U = \mathbb{Q}$ and $t = 1$.
Does the exponential “converge” for function inputs?
The exponential series $\sum_{n = 0}^\infty \frac{x^n}{n!}$ converges for every $x \in \mathbb{C}$. In particular, If we have some function $f : D \to \mathbb{C}$ and some $t \in D$, $\sum_{n = 0}^\infty \frac{f(t)^n}{n!}$ converges.
I mean it certainly approaches one single infinite series (polynomial for polynomial inputs) but do we consider that convergence?
From what you wrote I think you are asking whether the resulting function will be an analytic function? This will certainly be the case if $f$ itself is an analytic function, since analytic functions are closed under composition.
Also, is there anything useful you can take away from taking the exponential of a function?
I cannot really answer this part, but there are definitly cases where this is done, e.g. when solving a linear ode of first order with variable coefficients.
