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(Remark: I've edited the original answer to match the change in terminology, and I added some new comments.)

This is a theorem of Rosenstein from the paper $\aleph_0$-categoricity of groups. But note that he uses the terminology bounded order rather than finite exponent.

Theorem. (Rosenstein 1971)
An infinite abelian group is $\aleph_0$-categorical if and only if it has finite exponents.

Theorem 1. An $\aleph_0$-categorical group has finite exponents.

Theorem 2. An infinite abelian group of finite exponents is $\aleph_0$-categorical.

See Rosenstein's paper. The main tool is a structure theorem for an abelian group of finite exponents as direct sums of cyclic groups. This is apparently called Prufer's First Theorem.

1. By YCor's comment about $\mathbb{Q}^{(\omega)}$, one cannot prove Theorem 1 by considering only orbits of singletons. ($\mathbb{Q}$ works also.)

2. It is easier to prove that an $\aleph_0$-categorical torsion group $G$ has finite exponents. Indeed, if not then by compactness/DLS there is a countable model of $\operatorname{Th}(G)$ with an element of infinite order, which cannot be $G$. But if $G$ is not a torsion group, I don't see a quick way to avoid Ryll-Nardzewski or Omitting Types of some kind (although one can substitute other facts that use these results, e.g., in an $\aleph_0$-categorical structure, the algebraic closure of a finite set is finite).

3. An $\aleph_0$-categorical group need not be abelian. An example is the countably infinite extraspecial $p$-group (see Definition 5.15 here). On the other hand, there are many results in model theory along the lines of "$\aleph_0$-categorical plus some model-theoretic property" implies some abelian-like structure. For example, Bauer, Cherlin, and Macintyre showed that an $\aleph_0$-categorical superstable group is abelian-by-finite.

4. An infinite group of finite exponent need not be $\aleph_0$-categorical. For example, take an infinite finitely generated group of finite exponents (like a Tarski monster). Indeed, by the more general result about algebraic closure stated above, if $G$ is $\aleph_0$-categorical then any finite subset of $G$ generates a finite subgroup.

This is a theorem of Rosenstein from the paper $\aleph_0$-categoricity of groups. But note that he uses the terminology bounded order rather than finite exponent.

Theorem. (Rosenstein 1971)
An infinite abelian group is $\aleph_0$-categorical if and only if it has finite exponent.

Theorem 1. An $\aleph_0$-categorical group has finite exponent.

Theorem 2. An infinite abelian group of finite exponent is $\aleph_0$-categorical.

See Rosenstein's paper. The main tool is a structure theorem for abelian groups of finite exponent as direct sums of cyclic groups. This is apparently called Prufer's First Theorem.

1. By YCor's comment about $\mathbb{Q}^{(\omega)}$, one cannot prove Theorem 1 by considering only orbits of singletons. ($\mathbb{Q}$ works also.)

2. It is easier to prove that an $\aleph_0$-categorical torsion group $G$ has finite exponent. Indeed, if not then by compactness/DLS there is a countable model of $\operatorname{Th}(G)$ with an element of infinite order, which cannot be $G$. But if $G$ is not a torsion group, I don't see a quick way to avoid Ryll-Nardzewski or Omitting Types of some kind (although one can substitute other facts that use these results, e.g., in an $\aleph_0$-categorical structure, the algebraic closure of a finite set is finite).

3. An $\aleph_0$-categorical group need not be abelian. An example is the countably infinite extraspecial $p$-group (see Definition 5.15 here). On the other hand, there are many results in model theory along the lines of "$\aleph_0$-categorical plus some model-theoretic property" implies some abelian-like structure. For example, Bauer, Cherlin, and Macintyre showed that an $\aleph_0$-categorical superstable group is abelian-by-finite.

4. An infinite group of finite exponent need not be $\aleph_0$-categorical. For example, take an infinite finitely generated group of finite exponent (like a Tarski monster). Indeed, by the more general result about algebraic closure stated above, if $G$ is $\aleph_0$-categorical then any finite subset of $G$ generates a finite subgroup.

Theorem. (Rosenstein 1971)
An infinite abelian group is $\aleph_0$-categorical if and only if it has finite exponent.

Appendix. The discussion below has inspired me to add some discussion of the proof of theorems, and some other remarks. The Theorem above comes from two results in Rosenstein's paper.

Theorem 1. An $\aleph_0$-categorical group is has finite exponent.

Proof. Rosenstein uses the followong fact, due to due to Ryll-Nardzewski/Enegler/Svenonius independently.

Theorem 2. An infinite abelian group of finite exponent is $\aleph_0$-categorical.

See Rosenstein's paper. The main tool is a structure theorem for abelian group of finite exponent as direct sums of cyclic groups. This is apparently called Prufer's First Theorem.

1. By YCor's comment about $\mathbb{Q}^{(\omega)}$, one cannot prove Theorem 1 by considering only orbits of singletons. ($\mathbb{Q}$ works also.)

2. It is easier to prove that an $\aleph_0$-categorical torsion group $G$ has finite exponent. Indeed, if not then by compactness/DLS there is a countable model of $\operatorname{Th}(G)$ with an element of infinite order, which cannot be $G$. But if $G$ is not a torsion group, I don't see a quick way to avoid Ryll-Nardzewski or Omitting Types of some kind (although one can substitute other facts that use these results, e.g., in an $\aleph_0$-categorical structure, the algebraic closure of a finite set is finite).

3. An $\aleph_0$-categorical group need not be abelian. An example is the countably infinite extraspecial $p$-group (see Definition 5.15 here). On the other hand, there are many results in model theory along the lines of "$\aleph_0$-categorical plus some model theoretic property" implies some abelian-like structure. For example, Bauer, Cherlin, and Macintyre showed that an $\aleph_0$-categorical superstable group is abelian-by-finite.

4. An infinite group of finite exponent need not be $\aleph_0$-categorical. For example take an infinite finitely generated group of finite exponent (like a Tarski monster). Indeed, by the more general result about algebraic closure stated above, if $G$ is $\aleph_0$-categorical then any finite subset of $G$ generates a finite subgroup.

Theorem. (Rosenstein 1971)
An infinite abelian group is $\aleph_0$-categorical if and only if it has finite exponents.

Appendix. The discussion below has inspired me to add some discussion of the proof of theorems, and some other remarks. The theorem above comes from two results in Rosenstein's paper.

Theorem 1. An $\aleph_0$-categorical group has finite exponents.

Proof. Rosenstein uses the following fact, due to Ryll-Nardzewski/Enegler/Svenonius independently.

Theorem 2. An infinite abelian group of finite exponents is $\aleph_0$-categorical.

See Rosenstein's paper. The main tool is a structure theorem for an abelian group of finite exponents as direct sums of cyclic groups. This is apparently called Prufer's First Theorem.

1. By YCor's comment about $\mathbb{Q}^{(\omega)}$, one cannot prove Theorem 1 by considering only orbits of singletons. ($\mathbb{Q}$ works also.)

2. It is easier to prove that an $\aleph_0$-categorical torsion group $G$ has finite exponents. Indeed, if not then by compactness/DLS there is a countable model of $\operatorname{Th}(G)$ with an element of infinite order, which cannot be $G$. But if $G$ is not a torsion group, I don't see a quick way to avoid Ryll-Nardzewski or Omitting Types of some kind (although one can substitute other facts that use these results, e.g., in an $\aleph_0$-categorical structure, the algebraic closure of a finite set is finite).

3. An $\aleph_0$-categorical group need not be abelian. An example is the countably infinite extraspecial $p$-group (see Definition 5.15 here). On the other hand, there are many results in model theory along the lines of "$\aleph_0$-categorical plus some model-theoretic property" implies some abelian-like structure. For example, Bauer, Cherlin, and Macintyre showed that an $\aleph_0$-categorical superstable group is abelian-by-finite.

4. An infinite group of finite exponent need not be $\aleph_0$-categorical. For example, take an infinite finitely generated group of finite exponents (like a Tarski monster). Indeed, by the more general result about algebraic closure stated above, if $G$ is $\aleph_0$-categorical then any finite subset of $G$ generates a finite subgroup.

(Remark: I've edited original answer to match the change in terminology, and I added some new comments.)

Appendix. The discussion below has inspired me to add a sketch of the proof of the Theorem (from standard results), and some other remarks.

Theorem A. A countably infinite structure $M$ is $\aleph_0$-categorical if and only if for all $m>0$, the action of $\operatorname{Aut}(M)$ on $M^m$ has finitely many orbits.

Theorem B. Two countable abelian groups of finite exponent are isomorphic if and only if, for all $n>0$, they have the same number (possibly infinite) of elements of order $n$.

Theorem A is due to Ryll-Nardzewski/Enegler/Svenonius independently. I don't know the right attribution for Theorem B. One only needs to know that an abelian group of finite exponent is a direct sum of cyclic groups, which is apparently called Prufer's First Theorem.

Now the proofs (repeating my comment and YCor's comment below).

Theorem 1. An $\aleph_0$-categorical group is has finite exponent.

Proof. We use Theorem A. Suppose $\operatorname{Th}(G)$ is $\aleph_0$-categorical. We may assume $G$ is countable. Since elements of distinct orders are in distinct orbits of the action of $\operatorname{Aut}(G)$ on $G$, there is a uniform bound on the orders of torsion elements in $G$. Moreover, $G$ has no elements of infinite order since, if $g$ were such then $(g,g^n)$ for varying $n$ would be in distinct orbits in the action on $G^2$.

Proof. Suppose $G$ is an abelian group of finite exponent and $H$ is a countable group elementarily equivalent to $G$. Then $H$ is an abelian group of finite exponent having the same number of elements of every order as $G$ does, since all of this is in the first-order theory. So $G\cong H$ by Theorem B.

Some remarks on these proofs:

1. By another of YCor's comments (about $\mathbb{Q}^{(\omega)}$), one cannot prove Theorem 1 by considering only orbits of singletons. ($\mathbb{Q}$ works also.)

2. It is easier to prove that an $\aleph_0$-categorical torsion group $G$ has finite exponent. Indeed, if not then by compactness/LHS there is a countable model of $\operatorname{Th}(G)$ with an element of infinite order, which cannot be $G$. But if $G$ is not a torsion group, I don't see a quick way to avoid Ryll-Nardzewski or Omitting Types of some kind (although one can substitute other facts that use these results, e.g., in an $\aleph_0$-categorical structure, the algebraic closure of a finite set is finite).

3. The proof of Theorem 1 above is about the same as Rosenstein but he does say a little more. His proof of Theorem 2 takes up about a page and a half and it looks like this is because he is incorporating some of the content of the proof of Theorem B. (I did not read it very carefully though.)

Some final comments:

1. An $\aleph_0$-categorical group need not be abelian. An example is the countably infinite extraspecial $p$-group (see Definition 5.15 [here][2]). On the other hand, there are many results in model theory along the lines of "$\aleph_0$-categorical plus some model theoretic property" implies some abelian-like structure. For example, Bauer, Cherlin, and Macintyre showed that an $\aleph_0$-categorical superstable group is abelian-by-finite.

2. An infinite group of finite exponent need not be $\aleph_0$-categorical. For example take an infinite finitely generated group of finite exponent (like a Tarski monster). Indeed, by the more general result about algebraic closure stated above, if $G$ is $\aleph_0$-categorical then any finite subset of $G$ generates a finite subgroup.

(Remark: I've edited the original answer to match the change in terminology, and I added some new comments.)

Appendix. The discussion below has inspired me to add some discussion of the proof of theorems, and some other remarks. The Theorem above comes from two results in Rosenstein's paper.

Theorem 1. An $\aleph_0$-categorical group is has finite exponent.

Proof. Rosenstein uses the followong fact, due to due to Ryll-Nardzewski/Enegler/Svenonius independently.

A countably infinite structure $M$ is $\aleph_0$-categorical if and only if for all $m>0$, the action of $\operatorname{Aut}(M)$ on $M^m$ has finitely many orbits.

Now suppose $\operatorname{Th}(G)$ is $\aleph_0$-categorical. We may assume $G$ is countable. Since elements of distinct orders are in distinct orbits of the action of $\operatorname{Aut}(G)$ on $G$, there is a uniform bound on the orders of torsion elements in $G$. Moreover, $G$ has no elements of infinite order since, if $g$ were such then $(g,g^n)$ for varying $n$ would be in distinct orbits in the action on $G^2$.

See Rosenstein's paper. The main tool is a structure theorem for abelian group of finite exponent as direct sums of cyclic groups. This is apparently called Prufer's First Theorem.

Some remarks:

1. By YCor's comment about $\mathbb{Q}^{(\omega)}$, one cannot prove Theorem 1 by considering only orbits of singletons. ($\mathbb{Q}$ works also.)

2. It is easier to prove that an $\aleph_0$-categorical torsion group $G$ has finite exponent. Indeed, if not then by compactness/DLS there is a countable model of $\operatorname{Th}(G)$ with an element of infinite order, which cannot be $G$. But if $G$ is not a torsion group, I don't see a quick way to avoid Ryll-Nardzewski or Omitting Types of some kind (although one can substitute other facts that use these results, e.g., in an $\aleph_0$-categorical structure, the algebraic closure of a finite set is finite).

3. An $\aleph_0$-categorical group need not be abelian. An example is the countably infinite extraspecial $p$-group (see Definition 5.15 here). On the other hand, there are many results in model theory along the lines of "$\aleph_0$-categorical plus some model theoretic property" implies some abelian-like structure. For example, Bauer, Cherlin, and Macintyre showed that an $\aleph_0$-categorical superstable group is abelian-by-finite.

4. An infinite group of finite exponent need not be $\aleph_0$-categorical. For example take an infinite finitely generated group of finite exponent (like a Tarski monster). Indeed, by the more general result about algebraic closure stated above, if $G$ is $\aleph_0$-categorical then any finite subset of $G$ generates a finite subgroup.