You were close, but you missed the point of universality, so let me do the argument you had in mind, instead of a different one (the route chosen by tetrapharmakon). I agree that tetrapharmakon's argument is the efficient way to go, but I think it's a worthwhile exercise to do it without implicitly appealing to Yoneda. More to the point: Explicitly following the maps through all the natural isomorphisms in tetrapharmakon's answer is quite painful (at least to me), so a direct argument involving only the definitions makes me feel more comfortable.
Let me start from scratch (because I'm old and rigid, I'm unable to work with $F$ and $G$ in adjunctions, so please bear with me and let me replace them by the more descriptive letters $L$ and $R$).
So, we're given an adjunction $L : \mathscr{A} \longleftrightarrow \mathscr{C} : R$.
We're given $C,D \in \mathscr{C}$ and a product diagram $C\;\xleftarrow{p_C}\; C \times D\; \xrightarrow{p_D}\; D$. Applying the right adjoint $R$ we get the diagram $R(C)\; \xleftarrow{R(p_D)}\; R(C \times D) \;\xrightarrow{R(p_D)}\;R(D)$.
We want to see that this diagram is a product of $R(C)$ and $R(D)$. The only thing we need to do is to check the universal property.
So let us be given morphisms $R(C)\; \xleftarrow{a_C} \; A \; \xrightarrow{a_D} \; R(D)$ and we want to show that there is a unique morphism $d:A \to R(C \times D)$ such that $R(p_C)d = a_C$ and $R(p_D)d = a_D$. By adjointness $\operatorname{Hom}_{\mathscr{C}}(LA, C \times D) = \operatorname{Hom}_{\mathscr{A}}(A,R(C\times D)),$ so we can translate the problem into a problem in $\mathscr{C}$ by applying $L$. Thus, we consider the diagram $LR(C) \; \xleftarrow{L(a_C)}\; L(A) \;\xrightarrow{L(a_D)}\;LR(D)$. This is not quite where we want to be, but remembering the triangular identities for counit $\varepsilon: LR \Rightarrow 1_\mathscr{C}$ and unit $\eta:1_{\mathcal{A}} \Rightarrow RL$ leads us to $C \; \xleftarrow{\varepsilon_C L(a_C)}\; L(A)\;\xrightarrow{\varepsilon_D L(a_D)}\;D.$ Now, applying the universal property of the product diagram we started with, we finally find a unique morphism $e:L(A) \to C \times D$ such that $p_Ce = \varepsilon_C L(a_C)$ and $p_De= \varepsilon_D L(a_D)$.
Now the composition $d$ of $A \; \xrightarrow{\eta_{A}}\; RL(A)\;\xrightarrow{R(e)} \; R(C \times D)$ is the morphism we're looking for. Indeed, since $\eta : 1_{\mathcal{A}} \Rightarrow RL$ is a natural transformation, we have the commutative diagram
$\begin{array}{ccc} A & \xrightarrow{a_C} & R(C) \\ \downarrow{\scriptstyle \eta_A} & & \downarrow{\scriptstyle \eta_{RC}} \\ RL(A) & \xrightarrow{RL(a_C)} & RLR(C) \end{array}$
and combining this with the triangular identity $R(\varepsilon_C)\eta_{RC} = 1_{RC}$ we get
$R(p_C)d = R(p_C)R(e)\eta_A = R(p_Ce) \eta_A = R(\varepsilon_C L(a_C)) \eta_A = R(\varepsilon_C) RL(a_C) \eta_A = R(\varepsilon_C) \eta_{RC} a_C = a_C$
and similarly $R(p_D)d = a_D$. I leave it to you to convince yourself of the uniqueness of $d$.
Finally, let me stress that exactly the same argument works with general limits instead of binary products:
Given a diagram $D: \mathscr{D} \to \mathscr{C}$ with limit $C = \varprojlim_{\mathscr{D}} D$ (constant diagram) and universal morphism $u: C \Rightarrow D$, the morphism $R(u): R(C) \Rightarrow R \circ D$ exhibits $R(C)$ as $\varprojlim_{\mathscr{D}} R\circ D$. The only thing that needs to be checked is given any diagram $A: \mathscr{D} \to \mathscr{A}$ with morphism $a: A \Rightarrow R \circ D$, the morphism $a$ factors uniquely as $a = R(u)d$ for a morphism $d: A \Rightarrow R(C)$. The morphism $d$ is obtained as $R(e) \eta A$, where $e: L A \Rightarrow C$ is obtained from the universality of $u$ applied to $\varepsilon LA$.
As usual in category theory, this is all rather tautological (but admittedly confusing at first sight).