Non-standard tensor products of inner product spaces

Assuming complex scalars, no, any inner product which satisfies (2) for all $v \in \mathcal{V}$ and $w \in \mathcal{W}$ has the given form. To see this, let $[\cdot,\cdot]$ be any inner product which satisfies (2). So right away we know that $[v\otimes w,v\otimes w] = \langle v\otimes w, v\otimes w\rangle$ for all $v$ and $w$. Next, for $v,v' \in \mathcal{V}$ and $w \in \mathcal{W}$, we know that $$[(v + v')\otimes w, (v + v')\otimes w] = \langle (v + v')\otimes w, (v + v')\otimes w\rangle.$$ Expanding this out and applying $[v\otimes w,v\otimes w] = \langle v\otimes w,v\otimes w\rangle$ plus the same for $v'\otimes w$ yields $2{\rm Re}[v\otimes w, v'\otimes w] = 2{\rm Re}\langle v\otimes w, v'\otimes w\rangle$. Replacing $v$ with $iv$ yields equality of the imaginary parts too. So now we know that $[v\otimes w,v'\otimes w] = \langle v\otimes w,v'\otimes w\rangle$ for all $v$, $v'$, and $w$.

Finally, for any $v,v' \in \mathcal{V}$ and $w,w' \in \mathcal{W}$ we have $$[(v + v')\otimes (w + w'), (v + v')\otimes (w + w')] = \langle (v + v')\otimes (w + w'), (v + v')\otimes (w + w')\rangle.$$ Expanding this out and cancelling all the equalities we already have then leaves only $$2{\rm Re}([v\otimes w, v'\otimes w'] + [v\otimes w', v'\otimes w]) = 2{\rm Re}(\langle v\otimes w, v'\otimes w'\rangle + \langle v\otimes w', v'\otimes w\rangle).$$ Replacing $v$ with $iv$ yields equality of the imaginary parts too, so that $$[v\otimes w, v'\otimes w'] + [v\otimes w', v'\otimes w] = \langle v\otimes w, v'\otimes w'\rangle + \langle v\otimes w', v'\otimes w\rangle.$$ Then replacing $w'$ with $iw'$ yields $$[v\otimes w, v'\otimes w'] - [v\otimes w', v'\otimes w] = \langle v\otimes w, v'\otimes w'\rangle - \langle v\otimes w', v'\otimes w\rangle,$$ so that $[v\otimes w, v'\otimes w'] = \langle v\otimes w,v'\otimes w'\rangle$. As every element of the algebraic tensor product $\mathcal{V}\otimes\mathcal{W}$ is a linear combination of elementary tensors, this shows that $[\cdot,\cdot] = \langle\cdot,\cdot\rangle$.

I feel there ought to be a one-line proof of this, but I don't quite see it.

Let $V, W$ be 2-dimensional with orthonormal bases $\{v_0, v_1\}$ and $\{w_0, w_1\}$. If you expand out your condition (2), you obtain the following conditions: $$ \langle v_i \otimes w_j, v_i \otimes w_j \rangle = 1 \\ \langle v_i \otimes w_j, v_i \otimes w_k \rangle = \langle v_i \otimes w_j, v_k \otimes w_j \rangle = 0 \\ \langle v_0 \otimes w_0, v_1 \otimes w_1 \rangle + \langle v_1 \otimes w_0, v_0 \otimes w_1 \rangle = 0 $$ Clearly you can pick any value for $\langle v_0 \otimes w_0, v_1 \otimes w_1 \rangle$. Any nonzero choice will not satisfy your condition (1), since the RHS will be 0. So it is certainly possible. I'm not sure if they're of interest.

EDIT: Since there's disagreement in the comments, I'm posting my calculation. Consider general vectors $a_0 v_0 + a_1 v_1 \in V$, $b_0 w_0 + b_1 w_1 \in W$. Let's mangle Einstein notation by putting $\langle v_i \otimes w_j, v_k \otimes w_l\rangle = g_{ij}^{kl}$.

(2) says that $ \langle a_0 v_0 + a_1 v_1, a_0 v_0 + a_1 v_1 \rangle = (a_0^2 + a_1^2)(b_0^2 + b_1^2)$. Expanding out the LHS gives

$$ a_0^2 b_0^2 g_{00}^{00} + a_1^2 b_0^2 g_{10}^{10} + a_0^2 b_1^2 g_{01}^{01} + a_1^2 b_1^2 g_{11}^{11} + \\ 2\left(a_1 a_0 b_0^2 g_{10}^{00} + a_0^2 b_1 b_0 g_{01}^{00} + a_1^2 b_1 b_0 g^{10}_{11} + a_1 a_0 b_1^2 g^{01}_{11}\right) + \\ 4a_1 a_0 b_1 b_0 \left(g^{00}_{11} + g^{10}_{01} \right).$$

By choosing unit vectors, it is clear that $g^{ij}_{ij} = 1$. So the first line cancels with the RHS. By choosing a sum of two unit vectors, it is clear that $g^{ij}_{ik} = g^{ij}_{kj} = 0$, so the second line vanishes. We are left with $4a_1 a_0 b_1 b_0 (g^{00}_{11} + g^{10}_{01})$ = 0. If we choose our $g$ so that $g^{00}_{11} + g^{10}_{01} = 0$, this quantity will always vanish, and so (2) will hold for any pair of vectors.