Sorry, the question makes no sense to me. A few points, in no particular order:

1. We may consider models of a classical first-order theory $T$ in any Boolean category. So there's really no sense in contrasting model theory with category theory; the former studies models of first-order theories in $\mathbf{Set}$, while the latter gives you the tools to study first-order models in much more general contexts. I don't see the contrast; as others have said, these are apples and oranges.

2. If we're going to study a theory $T$, then morphisms between models of $T$ should be assumed to preserve everything that can be expressed in the language of $T$. For example, if $T$ is an algebraic theory, we should consider homomorphisms between $T$-models. If $T$ is a first-order theory, then we should consider elementary embeddings between $T$-models. In fact (although I admit to not having much knowledge in this area), I think that category theory usually chooses the morphisms for you, automatically. For example, if $L$ is an Lawvere theory and $X,Y : L \rightarrow \mathbf{C}$ are models of $L$ in a finite product category $\mathbf{C}$, then a homomorphism $\varphi : X \rightarrow Y$ is just a natural transformation $X \Rightarrow Y$. We don't really get much choice in the matter!

3. Let me expand on the above point a little. If $T$ is a first-order theory in the language of $\{\in\}$ and $X$ and $Y$ are $T$-models, then there is nothing natural about functions $f : X \rightarrow Y$ satisfying $x \in y \rightarrow f(x) \in f(y)$. Remember, $T$ is best viewed as a first-order theory, not as some particular presentation for that theory! The fact that we decided to axiomatize $T$ using the signature $\{\in\},$ rather than $\{\subseteq,x \mapsto \{x\}\}$ or something else entirely, is irrelevant, and in any reasonable definition of "first-order theory", the theory $T$ would not "remember" how it was defined. So we really do have to preserve all the structure that $T$ can express; we don't get a choice in the matter.

4. Let $\mathsf{AlgGrp}$ denote the algebraic theory of groups. Let $\mathsf{1stGrp}$ denote the 1st-order theory freely generated by $\mathsf{AlgGrp}$, whatever that means. Write $\mathbf{AlgGrp}$ and $\mathbf{1stGrp}$ for the corresponding categories of models in $\mathbf{Set}$. Then there is a forgetful functor $\mathbf{1stGrp} \rightarrow \mathbf{AlgGrp}.$ It is faithful and surjective on objects. But it is not full.

5. On the other hand, if $T$ is an algebraic theory and $F(T)$ denotes the coherent theory freely generated by $T$, then (someone correct me if I'm wrong here) the corresponding categories of models in $\mathbf{Set}$ should be equivalent.

(Sorry for bumping this to the top. I should have checked the date it was posted and just simply let it lie.)

Really, the question makes no sense to me. A few points, in no particular order:

1. We may consider models of a classical first-order theory $T$ in any Boolean category. So there's really no sense in contrasting model theory with category theory; the former studies models of first-order theories in $\mathbf{Set}$, while the latter gives you the tools to study first-order models in much more general contexts. I don't see the contrast; as others have said, these are apples and oranges.

2. If we're going to study a theory $T$, then morphisms between models of $T$ should be assumed to preserve everything that can be expressed in the language of $T$. For example, if $T$ is an algebraic theory, we should consider homomorphisms between $T$-models. If $T$ is a first-order theory, then we should consider elementary embeddings between $T$-models. In fact (although I admit to not having much knowledge in this area), I think that category theory usually chooses the morphisms for you, automatically. For example, if $L$ is an Lawvere theory and $X,Y : L \rightarrow \mathbf{C}$ are models of $L$ in a finite product category $\mathbf{C}$, then a homomorphism $\varphi : X \rightarrow Y$ is just a natural transformation $X \Rightarrow Y$. We don't really get much choice in the matter!

3. Let me expand on the above point a little. If $T$ is a first-order theory in the language of $\{\in\}$ and $X$ and $Y$ are $T$-models, then there is nothing natural about functions $f : X \rightarrow Y$ satisfying $x \in y \rightarrow f(x) \in f(y)$. Remember, $T$ is best viewed as a first-order theory, not as some particular presentation for that theory! The fact that we decided to axiomatize $T$ using the signature $\{\in\},$ rather than $\{\subseteq,x \mapsto \{x\}\}$ or something else entirely, is irrelevant, and in any reasonable definition of "first-order theory", the theory $T$ would not "remember" how it was defined. So we really do have to preserve all the structure that $T$ can express; we don't get a choice in the matter.

4. Let $\mathsf{AlgGrp}$ denote the algebraic theory of groups. Let $\mathsf{1stGrp}$ denote the 1st-order theory freely generated by $\mathsf{AlgGrp}$, whatever that means. Write $\mathbf{AlgGrp}$ and $\mathbf{1stGrp}$ for the corresponding categories of models in $\mathbf{Set}$. Then there is a forgetful functor $\mathbf{1stGrp} \rightarrow \mathbf{AlgGrp}.$ It is faithful and surjective on objects. But it is not full.

5. On the other hand, if $T$ is an algebraic theory and $F(T)$ denotes the coherent theory freely generated by $T$, then (someone correct me if I'm wrong here) the corresponding categories of models in $\mathbf{Set}$ should be equivalent.

6. Category theory isn't inherently restricted to studying well-behaved categories, in much the same way as order-theory isn't inherently restricted to studying lattice-orders. Its harder if your category is missing lots of limits and colimits, etc. - just like its harder to understand a poset thats missing most of its meets and joins - but it can be done. The fact that categories like $\mathbf{1stGrp}$ whose morphisms are elementary embeddings lack most limits and colimits doesn't make them impossible to study. Although of course, the kinds of results you can expect to prove are different. Your theorems will end up "feeling" much more model-theoretic than algebraic, but you can still use category theory to prove them.

7. Let me just reiterate that category theory is bigger than algebra. True, its most successful application domain is (currently) in algebra, but as it matures, it will begin to be used more-and-more frequently to study categories of models of much more expressive theories than algebraic theories.