Note that formulas using $\land$ and $\lor$ gates (and possibly the constants $0$ and $1$) are known as monotone.
The complexity of monotone formula equivalence depends on how complex formulas are allowed. Let me start with the related problem of implication (entailment) between two formulas $\phi\vdash\psi$, which is easier to classify. Note that $\phi$ and $\psi$ are equivalent iff $\phi\vdash\psi$ and $\psi\vdash\phi$.
Given monotone formulas $\phi$ and $\psi$, the problem of determining whether $\phi\vdash\psi$ is
in P for $\psi$ a CNF (even non-monotone). Indeed, we have $\phi\vdash\bigwedge_i\psi_i$ iff $\forall i\,(\phi\vdash\psi_i)$, and if $\psi_i$ is a clause, and $e_i$ the corresponding assignment that falsifies $\psi_i$ and makes all variables outside $\psi_i$ true, then
$$\phi\vdash\psi_i\iff e_i(\phi)=0.$$
For the right-to-left direction: if $e_i(\phi)=0$, and $e$ is any assignment such that $e(\psi_i)=0$, then $e\le e_i$, hence $e(\phi)=0$ using the monotonicity of $\phi$.
in P for $\phi$ a DNF (possibly non-monotone), by a dual argument.
coNP-complete for $\phi$ a 2-CNF and $\psi$ a 3-DNF (or dually: $\phi$ a 3-CNF and $\psi$ a 2-DNF), and a fortiori for any larger classes of formulas. Indeed, it is coNP-hard to check if a given 3-DNF is valid; we can write (in polynomial time) any 3-DNF in the form $\psi(\vec x,\neg\vec x)$ where $\psi$ is a monotone 3-DNF, and then
$$\vdash\psi(\vec x,\neg\vec x)\iff\bigwedge_i(x_i\lor y_i)\vdash\psi(\vec x,\vec y).$$
As above, the left-to-right direction follows from the monotonicity of $\psi$.
Now, back to equivalence. On the one hand, we have
$$\phi\equiv\psi\iff\phi\vdash\psi\text{ and }\psi\vdash\phi;$$
on the other hand,
$$\phi\vdash\psi\iff \phi\equiv(\phi\land\psi)\iff\psi\equiv(\phi\lor\psi).$$
Thus, we see that for monotone $\phi$ and $\psi$, the problem of determining if $\phi\equiv\psi$ is
in P for $\phi$ and $\psi$ CNF.
in P for $\phi$ and $\psi$ DNF.
coNP-complete for general monotone $\phi$ and $\psi$. It is enough to take $\phi$ a CNF, and $\psi$ a conjunction of a CNF and a DNF (which means $\psi$ can be written as a depth-3 formula with either $\bigwedge$ or $\bigvee$ as the top connective). Dually, it is enough to take $\phi$ a DNF, and $\psi$ a disjunction of a CNF and a DNF (whence again a depth-3 formula with either $\bigwedge$ or $\bigvee$ on top).
The only case left is the complexity of equivalence of a monotone CNF to a monotone DNF. This problem is interesting. As I learned in [4], it was proved to be solvable in quasipolynomial time $O(n^{\log n})$ by [3], and in coNP with only $O((\log n)^2)$ nondeterministic bits by [2,5]. Furthermore, when one of the formulas has bounded width, the problem is in P by [1,6], and even in L by [4].
References:
[1] T. Eiter, G. Gottlob: Identifying the minimal transversals of a hypergraph and related problems, SIAM Journal on Computing 24 (1995), no. 6, pp. 1278–1304, doi: 10.1137/S0097539793250299.
[2] T. Eiter, G. Gottlob, K. Makino: New results on monotone dualization and generating hypergraph transversals, SIAM Journal on Computing 32 (2003), no. 2, pp. 514–537, doi: 10.1137/S009753970240639X.
[3] M. L. Fredman, L. Khachiyan: On the complexity of dualization of monotone disjunctive normal forms, Journal of Algorithms 21 (1996), no. 3, pp. 618–628, doi: 10.1006/jagm.1996.0062.
[4] J. Goldsmith, M. Hagen, M. Mundhenk: Complexity of DNF minimization and isomorphism testing for monotone formulas, Information and Computation
206 (2008), no. 6, pp. 760–775, doi: 10.1016/j.ic.2008.03.002.
[5] D. J. Kavvadias, E. C. Stavropoulos: Monotone Boolean dualization is in $\mathrm{coNP}[\log^2n]$, Information Processing Letters 85 (2003), no. 1, pp. 1–6, doi: doi.org/10.1016/S0020-0190(02)00346-0.
[6] N. Mishra, L. Pitt: Generating all maximal independent sets of bounded-degree hypergraphs. In: Proceedings of the Tenth Annual Conference on Computational Learning Theory (COLT 1997), 1997, pp. 211–217, doi: 10.1145/267460.267500.
