Commuting right adjoints implies commuting left adjoints.

There are two results that piece together to give you your answer.

Theorem 1. If $\mathcal{C} \overset{F}{\underset{G}{\rightleftarrows}} \mathcal{D} \overset{K}{\underset{H}{\rightleftarrows}} \mathcal{E}$ are functors with $F \dashv G$ and $K \dashv H$, then $G \circ H \dashv K \circ F$.

Theorem 2. If $F \dashv G_1$ and $F \dashv G_2$, then $G_1 \cong G_2$; and if $F_1 \dashv G$ and $F_2 \dashv G$, then $F_1 \cong F_2$.

So assume that $R_3 \circ R_1 = R_4 \circ R_2$. Then $L_1 \circ L_3 \dashv R_3 \circ R_1$ by Theorem 1, and $L_2 \circ L_4 \dashv R_4 \circ R_2 = R_3 \circ R_1$ by Theorem 1, and so $L_1 \circ L_3 \cong L_2 \circ L_4$ by Theorem 2.

Hence the square of left adjoints commutes up to natural isomorphism.

Unfortunately you cannot expect to obtain a strong result than commutativity up to natural isomorphism, since adjoints themselves are only defined up to natural isomorphism.

Here's an example. The following square certainly commutes, where $\Delta : \mathbf{Set} \to \mathbf{Set} \times \mathbf{Set}$ is the diagonal functor, defined on objects by $A \mapsto (A,A)$. $$\require{amsCD} \begin{CD} \mathbf{Set} \times \mathbf{Set} @<{\Delta}<< \mathbf{Set} \\ @V{\Delta \times \mathrm{id}}VV @VV{\Delta}V \\ \mathbf{Set} \times \mathbf{Set} \times \mathbf{Set} @<<{\Delta \times \mathrm{id}}< \mathbf{Set} \times \mathbf{Set} \end{CD}$$

Let ${+} : \mathbf{Set} \times \mathbf{Set} \to \mathbf{Set}$ be the coproduct functor defined on objects by $${+}(A,B) \mapsto A+B = (A \times \{ 0 \}) \cup (B \times \{ 1 \})$$ and let $+' : \mathbf{Set} \times \mathbf{Set} \to \mathbf{Set}$ be the coproduct functor defined on objects by $${+'}(A,B) \mapsto B+A = (A \times \{ 1 \}) \cup (B \times \{ 0 \})$$ Then ${+} \dashv \Delta$ and ${+'} \dashv \Delta$, and ${+} \times \mathrm{id} \dashv \Delta \times \mathrm{id}$ and ${+'} \times \mathrm{id} \dashv \Delta \times \mathrm{id}$, and so we obtain the following diagram of left adjoints

$$\require{AMScd} \begin{CD} \mathbf{Set} \times \mathbf{Set} @>{+}>> \mathbf{Set} \\ @A{{+} \times \mathrm{id}}AA @AA{+'}A \\ \mathbf{Set} \times \mathbf{Set} \times \mathbf{Set} @>>{{+'} \times \mathrm{id}}> \mathbf{Set} \times \mathbf{Set} \end{CD}$$

This commutes up to natural isomorphism, since $(A+B)+C \cong (A+'B)+'C$ naturally in $A,B,C$, but it does not commute strictly.