Here are two examples where $E_k$-algebras show up in algebraic geometry, for $1<k<\infty$ (actually, just $k=2$):
The work of Gaitsgory and Lurie on Weil's conjecture for function fields. See here, here, and here, especially Lectures 21-24. As far as I understand it, the rough idea is that you want to prove a version of the Grothendieck-Lefschetz trace formula, but for a non-constructible sheaf on a certain poorly behaved algebraic prestack (the Ran space). They seem to circumvent this by developing some version of Verdier duality in this setting, that can be viewed as an algebro-geometric version of Koszul duality for $E_2$-algebras. I am not sure if the theory of $E_2$-algebras is strictly necessary for the proof, or just provides motivation.
The work of Toën and Vezzosi on Bloch's conductor conjecture, see arXiv:1605.08941 and arXiv:1710.05902. A main ingredient in their strategy is also a trace formula, but in the noncommutative setting (smooth proper dg-categories). It turns out that the dg-category $T$ they want to apply this to is not actually proper over the base. However, there exists a certain $E_2$-algebra $B$ over the base and $T$ turns out to be $B$-linear and proper as a $B$-linear dg-category. The trace formula they prove turns out to extend to the $E_2$-setting as well.
Regarding the comment about $E_1$-algebras: The derived category of a variety is equivalent to the derived category of modules over a certain $E_1$-algebra (the endomorphism algebra of the compact generator). A pretty active area of research in algebraic geometry is noncommutative algebraic geometry (via derived/dg-categories) and its relationship to birational geometry. See work of Bondal, Orlov, Kuznetsov, Bridgeland, Huybrechts, Van den Bergh, etc.
Edit: Here's another example.
- Arinkin-Gaitsgory, Singular support on coherent sheaves and the Geometric Langlands conjecture, Selecta 2015. They develop the theory of singular support in the setting of (ind)-coherent sheaves, on pretty general objects (derived Artin stacks), and use this to give a precise formulation of the categorical geometric Langlands conjecture. To that end they revisit some of the work of Benson-Iyengar-Krause on supports in triangulated categories. It seems that the correct setting for this theory turns out to be a dg-category/stable infinity-category equipped with an action of some $E_2$-algebra.
The answer is no in general, and I can't think of reasonable conditions under which it is yes. David's answer was about the "wrong" monoidal structure but it is still helpful : the free functor is analogous to extension of scalars along a map of bialgebras, and this "should not" preserve the underlying tensor product (defined as soon as you have some kind of bialgebra structure).
(In fact this analogy can be made precise by considering free operads, so that counterexamples yield counterexamples)
What we can say in general is that the forgetful functor is (strong) symmetric monoidal and that the left adjoint is therefore oplax monoidal.
Let's give an example to see that it is just not strong monoidal : say $O$ is the initial operad, and $P$ arbitrary (you can pick the commutative operad for concreteness). In this case, our functor is simply the free $P$-algebra functor $C \to Alg_P(C)$, whose "formula" is $X \mapsto \coprod_n (P(n)\otimes X^{\otimes n})_{h\Sigma_n}$.
Now the canonical map $F(X \otimes Y) \to F(X) \otimes F(Y)$ is the one that sends $(P(n) \otimes (X\otimes Y)^{\otimes n})_{h\Sigma_n}$ to $(P(n) \otimes X^{\otimes n})_{h\Sigma_n} \otimes (P(n)\otimes Y^{\otimes n})_{h\Sigma_n}$
It seems very rare for this to be an equivalence. In fact, given how both sides look, it seems very rare for the even exist an abstract equivalence.
Now one could ask for a lax symmetric monoidal structure instead. By some sort of yoga with adjunctions, this could not be compatible with the symmetric monoidal structure on the right adjoint (because of the above), and so it seems a bit unnatural but one might still ask.
It turns out one can also disprove that. Let me give an example : let $P$ be the associative operad, and again $O$ is initial. If there existed a lax symmetric monoidal structure, $Free(-)$ would send algebras to algebras in algebras i.e. $\mathbb E_2$-algebras (or, if we're dealing with $1$-categories as a special case, commutative algebras). I claim that this is not the case : the free associative algebra on any set with more than one element is non-commutative, even if that set admits a monoid structure.
Best Answer
For any symmetric monoidal $\infty$-category $\mathcal C$ and commutative algebra $A\in \mathcal C$ (you're supposed to think of $\mathcal C$ being $Cat_\infty$ itself, or maybe $Pr^L$; and $A= QCoh(Spec(k)) = Mod_k$), there is an equivalence: $CAlg(\mathcal C)_{A/}\simeq CAlg(Mod_A(\mathcal C))$, where the right hand side is to be interpreted as you expect if $\mathcal C$ has enough appropriate colimits, and can be made sense of operadically in general).
This in particular means that if you have a diagram $A\to B\to C$ of commutative algebras, then $B,C$ can be made into $A$-modules canonically (in fact into commutative $A$-algebras, but you can forget most of that structure), and the map $B\to C$ can be made $A$-linear canonically.
You can apply this to $QCoh(-)$ applied to $X\to Y\to Spec(k)$ and get a canonical $Mod_k$-linear structure on $f^*:QCoh(Y)\to QCoh(X)$. By naturality, this coincides with the approach you describe in (1).
Now, given a symmetric monoidal $\infty$-category $A$, and an $A$-linear functor $f: M\to N$, if $f$ admits a right adjoint $f^R$, then $f^R$ is canonically lax-$A$-linear - this follows from the theory of relative adjunctions in Higher Algebra, but concretely boils down to the existence of (coherent, natural, compatible etc.) maps $a\otimes f^R(n)\to f^R(a\otimes m)$, which are simply the usual projection maps, the mate of $f(a\otimes f^R(m))\cong a\otimes ff^R(m)\to a\otimes m$, where the first equivalence is the $A$-linearity of $f$, and the second map is the co-unit of the $f\dashv f^R$-adjunction. It then becomes a property for $f^R$ to be $A$-linear, namely the property that these projection maps be isomorphisms.
A cool thing to note is that this projection map is always an isomorphism when $a\in A$ is dualizable. So in your situation, $f_*$ is always $Perf(k)$-linear. If furthermore $f_*$ preserves colimits, then because $Mod_k$ is generated under colimits by $Perf(k)$, it follows that $f_*$ is $Mod_k$-linear; and this is essentially an "if and only if" (if $f_*$ is $Mod_k$-linear, one can prove that it preserves arbitrary coproducts and thus arbitrary colimits). This happens quite often, e.g. if $X,Y$ are qcqs, but presumably more generally too.
I'm not sure what $f^!$ is in an algebro-geometric context, but pesumably some similar game can be played there.