Yes, it makes sense. The order of an element $g$ in some group is the least positive integer $n$ such that $g^n = 1$ (the identity of the group), if any such $n$ exists. If there is no such $n$, then the order of $g$ is defined to be $\infty$.
As noted in the comment by @Travis, you can take a small permutation group to get an example. For instance, the permutation $(1,2,3,4)$ in the symmetric group $S_4$ of degree $4$ (all permutations of the set $\{1,2,3,4\}$) has order $4$. This is because
$$(1,2,3,4)^1 = (1,2,3,4)\neq 1,$$
$$(1,2,3,4)^2 = (1,3)(2,4)\neq 1,$$
$$(1,2,3,4)^3 = (1,4,3,2)\neq 1$$
and
$$(1,2,3,4)^4 = 1,$$
so $4$ is the smallest power of $(1,2,3,4)$ that yields the identity.
For the additive group $\mathbb{Z}$ of integers, every non-zero element has infinite order. (Of course, here, we use additive notation, so to calculate the order of $g\in\mathbb{Z}$, we are looking for the least positive integer $n$ such that $ng = 0$, if any. But, unless $g = 0$, there is no such $n$, so the order of $g$ is $\infty$.)
You see that it's an operational system, has a neutral element, and is invertible straight from the table, so the only thing left to check is associativity.
Here, we have
$(aa)c = 1c = c \neq b = ad = a(ac)$.
(The good news is that, if an operational system isn't associative, it usually doesn't take too long to find a counterexample.)
Best Answer
This is a good question. As you may have gathered from the comments, in the absence of any further information about the group, the answer is no, all you can do is to successively try higher powers.
But in most situations that arise in practice you do have extra information of one kind or another. For example, if your element is a matrix over a finite field, or even over the rational number or a number field, then the set of all possible orders of elements is known (at least in theory).
In such situations, you typically have a multiplicative upper bound for the order. In other words, you know a (possibly very large) integer $N$ such that the order of the element $g$ divides $N$. Then you can proceed as follows
For all primes $p$ dividing $N$, compute $g^{N/p}$. If $g^{N/p} = 1$ for some $p$, then replace $N$ by $N/p$ and start again - note that there will be a maximum of $\log n$ reductions of this kind. Otherwise, if $g^{N/p} \ne 1$ for all $p$, then the order of $g$ must be $N$.
Note also that computing $g^N$ can be accomplished on $O(\log n)$ group operations, by writing $N$ in binary $N = 2^{n_1} + \cdots + 2^{n_k}$, and then you can compute $g^N$ as $g^{2^{n_1}} \cdots g^{2^{n_k}}$.