Equivalence between existence of limits and products+equalizers, for derivators $\require{AMScd}\def\D{\mathbb{D}}\def\prepull{\vcenter{\lrcorner}}$
It is well known that for a category $\cal C$ the existence of finite limits is equivalent to the existence of finite products and equalizers, or to the existence of a terminal object and pullbacks.
What about (pre)derivators? Is it true that, for example, the existence of finite homotopy limits (right Kan extensions), the existence of homotopy pushouts, and homotopy equalizers, can be interchanged?
More precisely, can one infer from the existence of all homotopy limits of shape $\prepull$, and from the presence of terminal objects, the existence of all homotopy limits of shape $\rightrightarrows$, and from this, in turn, the existence of all finite limits (homotopy limits indexed over a homotopy finite diagram)?
A rather naive attempt for such a a statement is: given $X\in\D(\prepull)$ there is a diagram $\bar X\in \D(\rightrightarrows)$ such that $t_{1,*}X\cong t_{2,*}\bar X$, if
$$
\begin{CD}
@. \D(\prepull)\\
@. @VVt_{1,*}V\\
\D(\rightrightarrows) @>>t_{2,*}> \D(e)
\end{CD}
$$ denote the respective homotopy limit functors. Conversely, given $\bar X$ there is $X$ satisfying the same condition.
I suspect this naive approach fails, though, especially because the proof of the fact that the existence of homotopy pushouts and finite homotopy colimits are equivalent appears here as 7.1, and relies on high-tech arguments. Moreover, this correspondence, if true and pushed to the highest coherence level, seems to set up an equivalence of derivators $\D^\prepull\cong \D^\rightrightarrows$, which is hard to believe.
 A: $\require{AMScd}\def\D{\mathbb{D}}\def\prepull{\vcenter{\lrcorner}}$
The result you mention from the paper by Ponto-Shulman does not say exactly what you stated. The setting is that of derivators (so you already have all the homotopy Kan extensions indexed by small cats in some fixed category of diagrams). In that setting they prove the following:
Theorem 7.1 (Ponto-Shulman). 
Let $A$ be a homotopy finite category. 


*

*If $\mathbb D$ is a derivator and $\mathbb E \subseteq \mathbb D(e)$ contains the initial object and is closed under [homotopy] pushouts, then it is closed under [homotopy] $A$-colimits;

*If $F \colon \mathbb D \to\mathbb E$ is a morphism of derivators that preserves the initial object and [homotopy]
pushouts, then it preserves [homotopy] $A$-colimits.


The idea of their proof is that you can actually construct homotopy $A$-colimits via homotopy pushouts. You are actually asking about the dual setting, which of course holds similarly. In what follows I'll show that in the dual of the above theorem one can add to the characterization of classes closed under taking homotopy limits via homotopy pullbacks, also a characterization that uses homotopy equalizers. The reason for that is simple: homotopy pullbacks can be computed via homotopy equalizers:
Consider the following categories:


*

*$Ob(\prepull)=\{0,1,2\}$, $\hom_\prepull(0,1)=\{a\}$, $\hom_\prepull(2,1)=\{b\}$;

*$Ob(\rightrightarrows)=\{0,1\}$, $\hom_{\rightrightarrows}(0,1)=\{a,b\}$;

*$Ob(\square)=\{-\infty,0,1,2\}$, $\hom_\square(0,1)=\{a\}$, $\hom_\square(2,1)=\{b\}$, $\hom_\square(-\infty,0)=\{c\}$, $\hom_\square(-\infty,2)=\{d\}$;

*$Ob(\to\rightrightarrows)=\{-\infty,0,1\}$, $\hom_{\to\rightrightarrows}(0,1)=\{a,b\}$, $\hom_{\to\rightrightarrows}(-\infty,0)=\{c\}$.


Consider also the following functors:


*

*$i_{\prepull}\colon\prepull\to \square$ is the obvious inclusion;

*$i_{\rightrightarrows}\colon\{\rightrightarrows\}\to \{\to\rightrightarrows\}$ is the obvious inclusion;

*$q_{\prepull}\colon \prepull\to \{\rightrightarrows\}$ is such that $q_\prepull(0)=0=q_\prepull(2)$, $q_\prepull(1)=1$;

*$q_{\square}\colon \square\to \{\to\rightrightarrows\}$ is such that $q_\square(-\infty)=-\infty$, $q_\square(0)=0=q_\square(2)$, $q_\square(1)=1$. (Of course, $q_\square(c)=c=q_\square(d)$, $q_\square(a)=a$ and $q_\square(b)=b$).


Given a derivator $\mathbb D$ and $X\in \mathbb D(\prepull)$, I claim that 
$$
(-\infty)^*(i_{\rightrightarrows})_*(q_\prepull)_*X\cong (-\infty)^*(i_\prepull)_*X.
$$
In other words, the object $\bar X$ you are asking for in your question is $(q_\prepull)_*X$. To prove the above claim you have to notice that $q_{\square} i_{\prepull}=i_{\rightrightarrows}q_{\prepull}$, hence $(q_{\square})_* (i_{\prepull})_*\cong(q_{\square} i_{\prepull})_*=(i_{\rightrightarrows}q_{\prepull})_*\cong (i_{\rightrightarrows})_*(q_{\prepull})_*$. Thus, we are reduced to verify that $(-\infty)^*(q_{\square})_*\cong (-\infty)^*$. Let $Y\in \mathbb D(\square)$, to compute $(-\infty)^*(q_{\square})_*Y$ one can use the axiom (Der.4) (i.e., Kan extensions are pointwise in a derivator), so that
$$
(-\infty)^*(q_{\square})_*Y\cong \mathrm{Hocolim}_{-\infty/q_{\square}} \mathrm{pr}^*Y
$$
where $\mathrm{pr}\colon -\infty/q_{\square}\to \square$ is the obvious projection. So let us describe the category $-\infty/q_{\square}$:


*

*$Ob(-\infty/q_{\square})=\{(-\infty,id),(-\infty,c),(-\infty,ac),(-\infty,bc)\}$

*the object $(-\infty,id)$ is clearly initial.
Hence, 
$$
\mathrm{Hocolim}_{-\infty/q_{\square}} \mathrm{pr}^*Y\cong (-\infty,id)^*\mathrm{pr}^*Y\cong (-\infty)^*Y,
$$
where the first isomorphism holds by the well-known fact that taking the homotopy limit indexed by a category with an initial object is the same as evaluating at the initial object. This concludes the proof.
Let me add a final remark. In the comments to the question, Marc Hoyois suggests you to look at Proposition 4.4.3.2 in Lurie's Higher Topos Theory. If you look inside the proof of that result (dualizing to your setting), you will see that it actually gives a proof that homotopy pullbacks can be constructed using equalizers in the setting of $(\infty,1)$-categories. In fact, if you follow the steps in that proof (adapted to the setting of derivators), you start with an object $X\in \mathbb D(\prepull)$ and construct a "incoherent" diagram in $\mathbb D(e)^{\rightrightarrows}$ of the form:
$$
X_0\times X_1\rightrightarrows X_2
$$
This is exactly the underlying diagram of the object $(q_{\prepull})_*X$ constructed above.

Edit:
Let me be very precise on the exact statement of the theorem one gets with the above argument:
Theorem. 
Let $A$ be a homotopy finite category.  If $F \colon \mathbb D \to\mathbb E$ is a morphism of derivators that preserves finite products and homotopy equalizers, then it preserves homotopy $A$-colimits.
Proof. By Ponto-Shulman it is enough to verify that $F$ preserves pullbacks, that is, given $X\in \D(\prepull)$, there is a natural isomorphism
$$
F((-\infty)^*(i_{\prepull})_*X)\cong (-\infty)^*(i_{\prepull})_*F(X).
$$
Now, we know that $(-\infty)^*(i_{\prepull})_*X\cong (-\infty)^*(i_{\rightrightarrows})_*(q_\prepull)_*X$ and, also using that $F$ commutes with $(-\infty)^*i_{\rightrightarrows}$, we get
\begin{align*}
F((-\infty)^*(i_{\prepull})_*X)&\cong F((-\infty)^*(i_{\rightrightarrows})_*(q_\prepull)_*X)\\
&\cong (-\infty)^*(i_{\rightrightarrows})_*F((q_\prepull)_*X)
\end{align*}
To conclude it is enough to show that the canonical morphism
$$
F(q_\prepull)_*X\to (q_\prepull)_*F(X)
$$
is an isomorphism. Since, in a derivator, a morphism is an iso if and only if its components are isos (this is usually the axiom (Der.2)), it is enough to look at the underlying diagrams. With some standard computation with the axiom (Der.4), one gets that the underlying diagram of $F(q_\prepull)_*X$ is $(F(X_0\times X_1)\rightrightarrows F(X_2))$, while the underlying diagram of $(q_\prepull)_*F(X)$ is $(F(X_0)\times F(X_1)\rightrightarrows F(X_2))$. Since $F$ also commutes with finite products, these diagrams are isomorphic, concluding our proof.\\\
