Sorry about the long delay. The answer is no. Take a regular CW subdivision of a connected non-orientable surface $X$, such that the intersection of any two faces is a face of both. For example, $\mathbb{RP}^2$ can be triangulated with $6$ vertices, $15$ edges and $10$ triangles by taking an icosahedron and quotienting by the antipodal map. Let $P$ be the poset of faces of this subdivision, plus an additional minimal element $\hat{0}$ and maximal element $\hat{1}$. I claim that $P$ has all required properties, but the quotients of $KP$ by $\langle p-q \rangle$ and $\langle p+q \rangle$ (with $p$ and $q$ the two sides of a diamond) are not isomorphic when $K$ is any field of characteristic not $2$.

**$P$ is a lattice:** The intersection of any two faces is another face, or empty. The corresponding meet is that intersection, or $\hat{0}$. And, of course, the meet of any $x$ with $\hat{1}$ is $x$.

**$P$ is graded:** By dimension of the face.

**$P$ is strongly connected (sketch):** Given any two chains $(v,e,f)$ and $(v',e',f')$ of vertex in edge in face, we want to show that we can change one to the other by changing one element of the chain at a time. Choose a path $v_0$, $v_1$, ..., $v_N$ from $v$ to $v'$, and let $e_i$ be the edge between $v_i$ and $v_{i+1}$. We can swivel $(v,e,f)$ around the vertex $v=v_0$ to be $(v_0, e_0, f_0)$ for some $f_0$ containing $e_0$. Then we can switch to $(v_1, e_0, f_0)$. Then we can swivel around to $(v_1, e_1, f_1)$ for some $f_1$ etcetera. Finally, we reach $(v_N, e_{N-1}, f_{N-1})$, and swiveling around $v_N=v'$ brings us home to $(v',e',f')$.

**$P$ is thin:** Consider any length $2$ interval $[x,y]$. If $x = \hat{0}$, then this is the fact that an edge has two endpoints. If $x$ is a vertex and $y$ a face, this is the fact that the vertex of a polygon is in two edges. If $y=\hat{1}$, this is the fact that every edge is in two faces.

**It is impossible to assign the edges of $P$ weights of $\pm 1$ so that product of the weights around every diamond is $-1$:** Suppose we had such an assignment. Consider any edge $e$, contained in faces $f_1$ and $f_2$, and containing vertices $v_1$ and $v_2$. In the diamond $\hat{0} < v_1, v_2 < e$, one of the sides has weight $1$ and the other has weight $-1$: Say $(\hat{0}, v_1, e)$ has weight $1$. Similarly, suppose that $(e, f_1, \hat{1})$ has weight $1$ and $(e, f_2, \hat{1})$ has weight $-1$. Draw tangent vectors to $X$ at the midpoint of $e$, with the first vector pointing in direction of $v_1$ and the second pointing into $f_1$. Their wedge gives an orientation of $X$ at the midpoint of $e$. The condition on diamonds of the form $v < e_1, e_2 < f$ implies that this orientation is consistent across all of $X$. This contradicts that $X$ is nonorientable.

**The more general nonisomrophism** What about a stranger isomorphism between the algebras? The OP confirms below that he already knows that such an isomorphism must send arrows $x \overset{a}{\longrightarrow} y$ to $\epsilon(x \to y) \cdot a$ for some $\epsilon(x \to y) \in K^{\ast}$. I claim this is also impossible. To simplify exposition, I'll only do this for the specific triangulation of $\mathbb{RP}^2$ mentioned above.

Consider a Mobius strip embedded in $X$. To be concrete, take the equatorial belt of the icosahedron. It's image in $\mathbb{RP}^2$ is a simplicial complex with $5$ vertices $v_1$, $v_2$, $v_3$, $v_4$, $v_5$. (All subscripts will be periodic modulo $5$.) There are $10$ edges: Let $e_i$ connect $v_i$ to $v_{i+1}$ and $e'_i$ connect $v_i$ to $v_{i+2}$. There are $5$ triangles: Let $t_i$ have vertices $(v_i, v_{i+1}, v_{i+2})$. Now, look at the product
$$\prod_{i=1}^5 \frac{\epsilon(\hat{0} \to v_i) \epsilon(v_i \to e_i)}{\epsilon(\hat{0} \to v_{i+1}) \epsilon(v_{i+1} \to e_i)} \prod_{i=1}^5 \frac{\epsilon(v_i \to e_{i-1}) \epsilon(e_{i-1} \to f_{i-1})}{\epsilon(v_i \to e_i) \epsilon(e_i \to f_{i-1})}\prod_{i=1}^5 \frac{\epsilon(e_i \to f_{i-1}) \epsilon(f_{i-1} \to \hat{1})}{\epsilon(e_i \to f_{i}) \epsilon(f_{i} \to \hat{1})}.$$
The whole product telescopes to $1$. But each fraction is the ratio of two sides of a diamond, and there are $15$ terms, so the product is also $(-1)^{15} = -1$.

I have a much messier version of this formula which works for any cellular division of a Mobius strip, but one is enough for the counterexample.

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