Let $R$ be an arbitrary ring (i.e., may not have unity) with characteristic $n\in\mathbb{N}$ (positive integer) i.e any element added to itself $n$ times always yields additive identity $0$.
Show that for any $k\mid n$, ring $R$ contains an element of order $k$.
If $km = n$, then for any $r\in R$ we have $kmr = k(mr) = 0$. So if we assume the order of $mr\in R$ to be $k'$ and assume also $k'< k$, then:
$$k'(mr) = (k'm)r = 0 $$
which Would contradict the choice of $n$, BUT, we don't have it established there necessarely exists an element of order $n$. (Such would exist, if $R$ was ring with identity).
It would seem obvious, since for all $a\in R$ we have $na = 0$, there has to be an element whose order is $n$, but is there?
Characteristic only provides that for every $a\in R$ $na = 0$, it doesn't exclude the possibility that for every $a\in R$ there exists $n'<n$ for which $n'a=0$. There is no contradiction.
We do have that the order of every element divides $n$. If $\mbox{ord} (a)=n'$, then $n'k=n$, though, not any closer to establishing the existence of $n$-th order element :<
Best Answer
It is true that in a ring without unity, of "characteristic" $n$, there must be an element of order $n$.
First, note: for a ring without unity, we must define the characteristic to be the minimum $n$ such that $nr = r$ for all $r$. (Equivalently, it's the LCM of the order of $r$ over all $r$ in the ring.) See here. Anyway this looks like the definition you're using, except you seem to have forgotten about the minimum part, which is critical if we want to conclude there's an element of order $n$.
Let's forget about the multiplication structure and just consider the underlying abelian group. Here is our theorem:
Theorem. Let $A$ be an abelian group. If $N = \text{LCM}(\{\text{order}(a) \mid a \in A\})$, then there exists an element $a \in A$ of order $N$.
Proof. First we show that if $a, b \in A$ have orders $m$ and $n$, and $m, n$ are relatively prime, then there is an element of order $mn$. We can just take $a + b$. Notice that $mn(a+b) = 0$, so the order divides $mn$. If the order were smaller, WLOG we have $m'n(a+b) = 0$ where $m' < m$. Now $(m'n)b = m'(nb) = 0$, so $(m'n)a = 0$, and since the order of $a$ is $m$, $m \mid m'n$. Since $m, n$ are relatively prime, $m \mid m'$, contradiction.
Now we can build an element of order $N$. Factorize $N$ as a product of prime powers: $N = p_1^{x_1} p_2^{x_2} \ldots p_k^{x_k}$. There is an element of order $p_i^{x_i}$ for each $i$: certainly there must be some element $r$ of order divisible by $p_i^{x_i}$, or $N$ would not be the LCM of all the orders, but then an appropriate multiple of $r$ has order $p_i^{x_i}$ exactly. So we get elements $r_1, r_2, \ldots, r_k$ whose orders are $p_i^{x_i}$. Now we may apply the above result repeatedly (since the orders are relatively prime) to get an element of order $N$ -- in fact, it's just $r_1 + r_2 + \cdots + r_k$.