Equivalences creating limits 
Do non-adjoint equivalences create limits?

In the course of trying to prove that equivalences create limits, the computations boil down to the triangle identities for adjunctions when taking the naive route (similar to what happens when trying to prove that equivalences are full). In that situation we can get around the need for the triangle identities by using the fact that equivalences are faithful, which doesn't require the triangle identities; a similar tactic may be possible here, but the naive approach has left me down in the weeds on a non-obvious diagram chase.
In detail, suppose $F:\mathcal{C}\simeq\mathcal{D}$ is an equivalence and $S:\mathcal{I}\to\mathcal{C}$ is a diagram of shape $\mathcal{I}$ in $\mathcal{C}$ such that $F\circ S$ has a limit $$\big(\varprojlim F\circ S,\{\pi_I\}_{I\in\mathcal{I}}:\Delta\varprojlim F\circ S\Rightarrow F\circ S\big)$$ in $\mathcal{D}$. Let $F^{\sim1}:\mathcal{D}\simeq\mathcal{C}$ be a pseudo-inverse of $F$, with $\eta:1_\mathcal{C}\cong F^{\sim1}\circ F$ and $\epsilon:F\circ F^{\sim1}\cong1_\mathcal{D}$. We assert that $$\big(F^{\sim1}(\varprojlim F\circ S),\{\eta^{-1}_{S(I)}\circ F^{\sim1}(\pi_I)\}_{I\in\mathcal{I}}:\Delta F^{\sim1}(\varprojlim F\circ S)\Rightarrow S\big)$$ is a limit of $S$ in $\mathcal{C}$. Indeed, it is a cone to $S$ since for all $f:I\to I'\in\mathcal{I}$

commutes in $\mathcal{C}$ by naturality of $\eta$ and naturality of $\{\pi_I\}_{I\in\mathcal{I}}$ under the image of $F^{\sim1}$. This cone is further terminal; suppose $$\big(L,\{\pi'_I\}_{I\in\mathcal{I}}:\Delta L\Rightarrow S\big)$$ is another cone to $S$ in $\mathcal{C}$. Then $$\big(F(L),\{F(\pi'_I)\}_{I\in\mathcal{I}}:\Delta F(L)\Rightarrow F\circ S\Big)$$ is a cone to $F\circ S$ in $\mathcal{D}$ and thusly induces a unique morphism of cones $$\langle F(\pi'_I)\rangle_{I\in\mathcal{I}}:F(L)\to\varprojlim F\circ S.$$ We then have that $$F^{\sim1}(\langle F(\pi'_I)\rangle_{I\in\mathcal{I}})\circ\eta_L:L\to F^{\sim1}(F(L))\to F^{\sim1}(\varprojlim F\circ S)$$ is a morphism of cones in $\mathcal{C}$ since

commutes for all $I\in\mathcal{I}$ by virtue of the following straightforward computations
$$\eta^{-1}_{S(I)}\circ F^{\sim1}(\pi_I)\circ F^{\sim1}(\langle F(\pi'_I)\rangle_{I\in\mathcal{I}})\circ\eta_L=\eta^{-1}_{S(I)}\circ F^{\sim1}(\pi_I\circ\langle F(\pi'_I)\rangle_{I\in\mathcal{I}})\circ\eta_L$$
$$=\eta^{-1}_{S(I)}\circ F^{\sim1}(F(\pi'_I))\circ\eta_L=\pi'_I\circ\eta_L^{-1}\circ\eta_L=\pi'_I.$$
This morphism of cones is further unique, since for any other arrow $$u:L\to F^{\sim1}(\varprojlim F\circ S)$$ satisfying $$\eta^{-1}_{S(I)}\circ F^{\sim1}(\pi_I)\circ u=\pi'_I\implies F^{\sim1}(\pi_I)\circ u=\eta_{S(I)}\circ\pi'_I$$ we have that $$\epsilon_{\varprojlim F\circ S}\circ F(u):F(L)\to F(F^{\sim1}(\varprojlim F\circ S))\to\varprojlim F\circ S$$ is a morphism of cones from $\big(F(L),\{F(\pi'_I\}_{I\in I}\big)$ to $\big(\varprojlim F\circ S,\{\pi_I\}_{I\in \mathcal{I}}\big)$ since
$$\pi_I\circ\epsilon_{\varprojlim F\circ S}\circ F(u)=\epsilon_{F(S(I))}\circ F(F^{\sim1}(\pi_I))\circ F(u)$$
$$=\epsilon_{F(S(I))}\circ F(F^{\sim1}(\pi_I)\circ u)=\epsilon_{F(S(I))}\circ F(\eta_{S(I)}\circ\pi'_I)=\epsilon_{F(S(I))}\circ F(\eta_{S(I)})\circ F(\pi'_I),$$ and this is where I'm stuck. If we have the triangle identities then $\epsilon_{F(S(I))}\circ F(\eta_{S(I)})=1_{F(S(I))}$ and we're good to go, so the proof is easy to complete for adjoint equivalences, but without this identity I don't see how to proceed. I've tried taking things under the image of $F^{\sim1}$ and even under the image of $F$ again after that, but ended up back at the triangle identities after the second application so it's likely that only applying $F^{\sim1}$ is the ticket if that approach is going to work. Any assistance is appreciated.

The answer to the question as (implicitly) posed might be 'no' because I'm using the following 'wrong' definition of creation of limits, where the reflecting part reflects existence, per the discussion here.

Definition. Let $S:\mathcal{I}\to\mathcal{C}$ be a diagram of shape $\mathcal{I}$ in $\mathcal{C}$. A functor $F:\mathcal{C}\to\mathcal{D}$ reflects the limit of $S$ iff $F\circ S$ having a limit in $\mathcal{D}$ means that $S$ has a limit in $\mathcal{C}$. Further, $F$ preserves the limit of $S$ iff $S$ has a limit in $\mathcal{C}$, $F\circ S$ has a limit in $\mathcal{D}$, and the canonical induced arrow from the image of the limit of $S$ in $\mathcal{C}$ under $F$ to the limit of $F\circ S$ in $\mathcal{D}$ is an isomorphism. We will say that $F$ creates the limit of $S$ iff $F$ reflects and preserves the limit of $S$.

If we use the version that doesn't reflect existence the proof might go through more smoothly, but then it is interesting that only adjoint equivalences 'create limits' in the above stronger sense.
 A: The obvious answer to the question in the title is that it depends on the definition of "creating limits"; but the argument you gave is trying to prove that an equivalence lifts limits up to isomorphism, let's restrict attention to that.
If you want to work at this level of attention to detail, it's better to treat the idea of "the limit $\lim S$ of the diagram $S$" as a fiction, and instead only think about cones, and the property of a cone being a limit cone.
What you want to prove is that given an equivalence $F : C \to D$ and a diagram $S : I \to C$, any limiting cone on $F \circ S$ is isomorphic to the image of some limiting cone on $S$ under $F$.
At this point we could appeal to some metatheorem and say that since this statement does not involve equality of objects of a category, it is equivalence-invariant, and therefore it suffices to check it when $F$ is the identity functor, in which case it is obvious.
If you want to give a hands-on proof, it makes things much easier to avoid ever thinking about the inverse functor $F^{\sim 1}$.
For instance, we could argue as follows.
Suppose given a limiting cone on $F \circ S$ with vertex $Y \in D$.
Since $F$ is essentially surjective, we can choose an object $X \in C$ and an isomorphism $FX \cong Y$.
By composing the original limiting cone with $FX \cong Y$, we obtain a new cone which is still limiting (check!), and now has vertex $FX$.
Now because $F$ is fully faithful, this cone is the image of a (unique) cone on $S$ with vertex $X$.
And this cone is limiting--we need to check that there is a unique map from any cone on $S$ to it, but this follows from the fact that $F$ is fully faithful plus the corresponding fact for the image cone.
This argument never considers things like $F(F^{\sim 1}(Y))$--that's where things get messy in a direct approach, because we would like to pretend this object is just $Y$ itself but we can't.
Instead, it's related to $Y$ by some additional data and then we have to worry about coherence of this data, etc.--much easier to avoid the whole situation in the first place.
