It is not true.
Consider the lattices $\mathbf{M}_3,\mathbf{M}_4, \dots$ only with the language $\{\vee,\wedge\}$ instead of $\{\vee,\wedge,\neg\}$. First of all, one can show that $\mathbf{M}_n$ is simple for $n\geq 3$. Second of all, since the class of all lattices is a congruence distibutive variety, we can use Jonsson's lemma to say that the subdirectly irreducibles in $\mathsf{V}(\mathbf{M}_n)$ belong to $\mathsf{HS}(\mathbf{M}_n)$. Since the only sublattices of $\mathbf{M}_n$ are the lattices $\mathbf{M}_k$ for $k=0,1,2,\dots,n$, and each of those (except $\mathbf{M}_1$ and $\mathbf{M}_2$) are simple, we get that the subdirectly irreducibles in $\mathsf{V}(\mathbf{M}_n)$ are exactly $\mathbf{M}_0,\mathbf{M}_3,\dots,\mathbf{M}_n$.
So $\mathsf{V}(\mathbf{M}_m)$ is strictly contained in $\mathsf{V}(\mathbf{M}_n)$ whenever $3\leq m<n$. So, by Birkhoff's theorem, there must be some universally quantified equation $\forall v, \phi(v)=\chi(v)$ satisfied by $\mathbf{M}_m$ that is not satisfied by $\mathbf{M}_n$. So in $\mathbf{M}_n$ $\exists v,\phi(v)\neq\chi(v)$ (if $\phi(v)<\chi(v)$, then switch $\phi$ and $\chi$). Then $\phi\models_{\mathbf{M}_m}\chi$, but $\phi\not\models_{\mathbf{M}_n}\chi$. Since $\phi,\chi$ are formulas in $\{\vee,\wedge\}$, they are also formulas in $\{\vee,\wedge,\neg\}$.
As a specific example, $\mathbf{M}_3$ satisfies the equation
$
x_1\vee((x_2\vee(x_3\wedge x_4))\wedge(x_3\vee(x_1\wedge x_4)))\approx x_1\vee(((x_1\vee x_2)\vee(x_3\wedge x_4))\wedge(x_3\wedge(x_2\vee x_4)))
$
but $\mathbf{M}_4$ does not. Let
$\chi(x_1,x_2,x_3,x_4)=x_1\vee((x_2\vee(x_3\wedge x_4))\wedge(x_3\vee(x_1\wedge x_4)))$
and
$\phi(x_1,x_2,x_3,x_4)=x_1\vee(((x_1\vee x_2)\vee(x_3\wedge x_4))\wedge(x_3\wedge(x_2\vee x_4)))$
Now $\phi\models_{\mathbf{}_3}\chi$ since $\phi(v)$ always returns the same element as $\chi(v)$, but $\phi\not\models_{\mathbf{M}_4}\chi$ since $\chi(1,2,3,4)=1$ and $\phi(1,2,3,4)=\top$.
Sometimes it helps to unpack the definition. In this case, for the forgetful functor $A/\mathcal{C} \to \mathcal{C}$ to have a left adjoint, you'd need to assign to each object $X$ an object $FX \in \mathcal{C}$ and a morphism $f_X : A \to FX$ such that
$$\mathrm{Hom}_{A/\mathcal{C}}(A \xrightarrow{f_X} FX, A \xrightarrow{g} Y) \cong \mathrm{Hom}_{\mathcal{C}}(X, Y)$$
naturally in $X \in \mathcal{C}$ and all $A \xrightarrow{g} Y \in A/\mathcal{C}$.
Can you see how you might define such a natural isomorphism? If not, hover over the box below for an additional hint.
Define $FX = A+X$ and $f_X = \iota_A : A \to A+X$, and note that given $g : A \to Y$, morphisms $h : A+X \to Y$ such that $h \circ i_A = g$ correspond naturally with morphisms $X \to Y$.
Best Answer
You probably know this already, but distributivity of $\vee$ over $\wedge$ in a lattice follows easily from distributivity of $\wedge$ over $\vee$: \begin{align*} (a\vee b)\wedge (a\vee c) &= ((a\vee b)\wedge a)\vee ((a\vee b)\wedge c)\\ &= a \vee ((a\wedge c)\vee (b\wedge c))\\ &= (a \vee (a \wedge c)) \vee (b \wedge c)\\ &= a \vee (b\wedge c). \end{align*} So if you know that $a\wedge -$ is a left adjoint, you actually get both distributive laws "for free".
To answer your actual question, $a\vee -$ is not always a right adjoint. For a counterexample, you just need to find a Heyting algebra in which $a\vee -$ fails to preserve some infinite meets. In his answer to this MathOverflow question, Andreas Blass gives a nice example: Consider the Heyting algebra of open subsets of $\mathbb{R}$. Let $a = \mathbb{R}\setminus \{0\}$, and let $b_n = (-\frac{1}{n},\frac{1}{n})$ for all $n\in \mathbb{N}^+$. Then $a\vee \left(\bigwedge_{n\in \mathbb{N}^+} b_n\right) = a$ but $\bigwedge_{n\in \mathbb{N}^+} (a\vee b_n) = \mathbb{R}$.
On the other hand, in a finite Heyting algebra, all meets are finite, so $a\lor -$ always preserves all meets. The existence of a left adjoint then follows by the adjoint functor theorem for posets. It is defined by $$L(x) = \bigwedge \{c\mid x\leq a\lor c\}.$$ Of course, verifying that this construction works requires you to know that $\vee$ distributes over $\wedge$. So this is not a route to proving the second distributive law, even in finite Heyting algebras.