Why are operads sometimes better than algebraic theories? Question 1: Are there any contexts in which replacing the category of (non-symmetric or symmetric) operads (in some monoidal category or symmetric monoidal category, respectively) with the category of their corresponding monads leads to undesirable behavior? (I know that for non-symmetric operads this functor is non-injective and so at least in one of the versions the question makes sense. Is it injective for symmetric operads?)
Question 2: Let $C$ be a monoidal category with projections and diagonal mappings that satisfies reasonably good conditions (a perfect example: an elementary topos with a Cartesian monoidal structure, but I'm targeting a much wider class). Are there contexts in which replacing algebraic theories defined by (symmetric/nonsymmetric) operads in this monoidal category with all algebraic theories in it leads to undesirable behavior? The motive of this question in general is: whenever the richness of the monoidal structure makes it possible to define a wider class of theories - are there reasons not to go to it? Thus, the transition from non-symmetric to symmetric operads is embedded in this motif when it comes to a symmetric monoidal category.
I read several sources for three months (and in parallel dealt with operads a little in one course at the university) and I have not yet developed a coherent vision of why people in all contexts do not write "algebraic monad"(in the sense of the monad of the algebraic theory that expressible in the considered monoidal category) instead of "operad", when they can.
P.S. I understand that there is not always a clear answer to this kind of question, but sometimes it is known and therefore its existence is part of the question.
 A: First - yes, for symmetric set-operads this functor is "injective", though it is not fully faithful. It is faithful on general maps and fully faithful on isomorphisms. Its image can easily be characterized: symmetric set-operads are the so-called "analytic monads" that is the monad whose underlying endofunctor preserves filtered colimits and connected limits, and whose structure map are weakly cartesian natural transformations. Morphisms of operads correspond to weakly cartesian morphisms of monads.
The first big difference (which is essentially what David White refers to) is of course that operads have their models in a symmetric monoidal categories, while Lawvere theories have their models in a cartesian monoidal categories - so indeed when you want to consider a model in chain complex etc (where the product is taken to be the tensor product, not the cartesian product) then you need an operad.
But if you are working with a monoidal category $C$ which is cartesian (that is the monoidal product is the cartesian product, which can be axiomatized in terms of the existence of projections and diagonal map satisfying some axiom), then, indeed, everything that can be axiomatized with an operads can be axiomatized by a Lawvere theory as well.
This being said I can cite two other situations where moving from operads to monads/Lawvere theories does "something wrong". The second one is in my opinion the most important one and honestly is the whole "raison d'être" of operades - both historically and in today's mathematics. But the first one might be more directly interesting if you are more of an algebraist than a homotopy theorist.

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*If $O$ is an operads and $A$ is a $O$-algebra, then there is a notion of $A$-module (maybe I should called then bimodule). The easiest way to see this is in the special case where $C$ is a monoidal category with coproduct preserved by the tensor product in each variables (but the notion makes sense more generally), then I can build a new monoidal category whose underlying category is $C^2$ but whose monoidal product is $(A,M) \otimes (A',M') = (A \otimes A', A \otimes M' \oplus A' \otimes M)$ and it makex sense to look at the model of our operads in this new monoidal category - which essentially corresponds to an $O$-algebra structure on $A$ and an $A$-module (or bimodule) structure on $M$. If $O$ is the commutative operad, this gives the usual notion of modules, if $O$ is the Associative operads this gives the usual notion of bimodules. Now, even if $C$ is cartesian, this monoidal structure on $C^2$ isn't so there is no analogue of this construction for Lawvere theories.


*When doing topology, the homotopy theory of operads behave very differently from the homotopy theory of Lawvere theories. In short, both operads and Lawvere theories have fairly natural notions of "weak models" (i.e. "models up to homotopy"), in both cases, they can be defined either directly, or in terms of some "cofibrant replacement" in a category of simplicial operads/simplicial Lawvere theory, but the weak algebras of an operad are very different (and much more interesting !) than the weak algebras of the corresponding Lawvere theory.
Let me give some details on this second point:
Homotopy models for Lawvere theories: If $C$ is a Lawvere theory (Which I see as a category with finite products) then I can define a notion of "homotopy model" or "weak model" of "C", by looking at (pseudo)-functor from $C$ to the $\infty$-category of spaces that preserves finite products (I mean send product in $C$ to homotopy products). Because Pseudo-functor to space spaces can always be strictified, this is the same as looking at actual functor $C \to Spaces$ (Where space is either topological space or simplicial sets) which sends products in $C$ to "homotopy products".
Now there is a "strictification" theorem for these: Any such "weak C-model" is homotopically equivalent to an actual model of $C$ in the category of spaces (either topological space or simplicial sets). In fact the model category of weak models and strict model are Quillen equivalent. As far as I know, this is originally due to Badzioch, in Algebraic theories in homotopy theory.
Also, another way to define "weak algebras" for a Lawvere theory is to work within the category of "simplicial Lawvere theory" which carries a model structure and defines the weak algebras as the algebras for some "cofibrant replacement" in the sense this model structure - but this ends-up being equivalent to the above.
Homotopy model of operads: There is also a definition of "homotopy algebras" for an operads $O$. It is a little harder to define without going into technical details - the simplest way to phrase in modern language it is to say that a set-operad is in particular an $\infty$-operad, and one can consider its models in the $\infty$-category of spaces in the sense of the theory of $\infty$-operads. But the notion was known long before $\infty$-category theory and is fairly natural from the point of view of topology: they corresponds to algebra (in the category of space/simplicial sets) for the Boardman-Vogt resolution (which essentially replace equation by homotopies in a coherent way), or more generally for any cofibrant resolution of your operad.
Now this time it is no longer true that every weak algebra is equivalent to a strict algebra in topological spaces or simplicial sets. This is only true if the operad is "$\Sigma$-cofibrant". For a set-operad this simply means that the actions of the symmetric groups are free - so, for example, the operads for commutative monoid isn't $\Sigma$-cofibrant.
I think this is due to Moerdijk and Berger here, but maybe this was know before?
Now, How is it different? Let's take the simplest example. The operads "Comm" for commutative monoids.
As an operads, its weak algebra are the space endowed with $E_\infty$ structure. So for example if we restrict to connected spaces (or "group-like" algebras) we get a homotopy theory equivalent to that of connective spectrum.
If we now see "Comm" as a lawvere theory, then its weak algebras are - because of Badzioch's theorem - homotopy equivalent to spaces endowed with a strictly commutative multiplication. These are much more restrictive than $E_\infty$-structures. I think (don't quote me on this) that if we restrict to connected (or group-like algebra) we get something equivalent to bounded chain complexes.
The difference is maybe easier to see if we look at models in groupoids or categories instead of spaces:

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*A weak model $X$ of Comm as an operad is a symmetric monoidal structure on $X$.

*A weak model $X$ of Comm as a Lawvere theory is the same as a "strict model" by (an appropriate variant of) Badzioch's theorem, that is the same as a groupoid with a "strictly commutative and associative" monoid structure.

The two are different exactly in the following way: Given an object, $x \in X$ in the first case, the $n$-fols products $x \otimes \dots \otimes x$ come equipped with an action of the symmetric group $\Sigma_n$ by permutation of the component. In the second case, there is no such action, to be more precise (in terms of weak model), this should be thought of as a symmetric monoidal category where the permutation action of $\Sigma_n$ is trivial on every power of every object - which you'll agree is something very uncommon for monoidal categories.
Edit: What about the "$\infty$-world" ?
So, that second point is really about how to go from $1$-operads to $\infty$-operads and from $1$-Lawvere theory to $\infty$-lawvere theories. It can be rephrased as the fact that the "square" that we want to draw whose corner are "$1$/$\infty$-operads/Lawvere theory" isn't a commutative square. So if one decide to focus on the $\infty$-world that problem does indeed disappear - or rather become the fact that the functor from $\infty$-operads to $\infty$-Lawvere theories doesn't preserves 1-truncated objects.
For example, the 1-Lawvere theory for commutative monoids comes from the 1-operads for commutative monoids (so the terminal symmetric operads), but when you see it as an $\infty$-Lawvere theory it is no longer the image of an operad: the operad for commutative monoids seen as an $\infty$-operads is sent to the $\infty$-Lawvere theory for $E_\infty$-algebras.
This does make the connection between $\infty$-operads and $\infty$-Lawvere theories "feel a bit different" than its 1-categorical counterparts, so I felt like it would be interesting to add a few comment about it:

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*First, the main difference remains: given a Lawvere theory you can only takes its models in a cartesian $\infty$-category, while you can look at the models of an operads in any monoidal $\infty$-category. So for example if you want to say something like "commutative ring spectra are commutative monoids in the category of spectra", that, as you are using the smash product of spectra which is not the cartesian product, then you really need the operads for commutative monoids and not the corresponding Lawvere theory (or to put it another way, you need to know that the Lawvere theory for $E_\infty$-algebras is "operadic").


*The discussion about modules for an $O$-algebras still applies completely unchanged to the $\infty$-world.


*Now, the way $\infty$-operads identifies with special monads/Lawvere theory become much cleaner in the $\infty$ setting: The $\infty$-category of $\infty$-operads identifies exactly with the $\infty$-category of "finitary polynomial monads" (and cartesian morphisms between them, and where finitary is interpreted to mean that their arities are finite sets, not finite spaces). The details of this have been worked out by Gepner, Haugseng and Kock in $\infty$-Operads as Analytic Monads.
A: Yes, there are many contexts where operads are better than Lawvere theories. For one thing, algebras over a Lawvere theory are defined in the category Set. If you want to study algebras in a different category (e.g., chain complexes) then Lawvere theories don't help you. A lot of applications of operads are in categories like chain complexes, where they help you study DGAs, CDGAs, Lie-algebras, Koszul duality, etc. Even more applications are in stable homotopy theory where we study operad-algebras in monoidal categories of spectra, encoding structured ring spectra. If you only had Lawvere theories and not operads, you really couldn't do chromatic homotopy theory.
Charles Rezk has a nice paper working out the analog of Lawvere theories for simplicial sets. But there is no analog that can act in spectra, as far as I know.
