The answer is both yes and no, so this will take a bit of elaboration.
There are two ways to define a ring. One of them require the existence of a $1$, the other does not.
Let's start with the one that does. In this case, a subring is required to contain the $1$ from the larger ring, and hence no proper ideal can be a subring (as any ideal containing $1$ will be the entire ring).
If we removed the requirement that the subring contained the original $1$, then the answer would be "sometimes", since the ideal might or might not have a "local" $1$ (I invite you to try some small examples of rings to find examples of either case).
If we instead take the case where a ring need not have a $1$, then it follows straight from the definitions that any ideal will also be a subring.
Here is a general way to get a proper non-trivial ideal which has a "local" $1$: Take any $x\in R$ such that $x^2 = x$ (such $x$ are called idempotent) and such that $x\neq 0$ and $x\neq 1$ (which need not exist, but will for example exist in any ring of the form $R_1\times R_2$). Then the ideal generated by $x$ will have $x$ as its "local" $1$ (I am assuming the ring to be commutative, or at least $x$ to be central here).
Conversely, if we have such an $x$ then the ring will be the direct product of the ideals generated by $x$ and $1-x$ (note that $1-x$ is also idempotent).