jmc

From $\mathbf{u}\times \mathbf{v}+\mathbf{u}=\mathbf{w}$ and $\mathbf{w}\times \mathbf{u}=\mathbf{v},$ $$(\mathbf{u}\times \mathbf{v}+\mathbf{u})\times \mathbf{u}=\mathbf{v}.$$Expanding, we get $$(\mathbf{u}\times \mathbf{v})\times \mathbf{u}+\mathbf{u}\times \mathbf{u}=\mathbf{v}.$$We know that $\mathbf{u}\times \mathbf{u}=0.$ By the vector triple product, for any vectors $\mathbf{p},$ $\mathbf{q},$ and $\mathbf{r},$ $$\mathbf{p}\times (\mathbf{q}\times \mathbf{r})=(\mathbf{p}\cdot \mathbf{r})\mathbf{q}-(\mathbf{p}\cdot \mathbf{q})\mathbf{r}.$$Hence, $$(\mathbf{u}\cdot \mathbf{u})\mathbf{v}-(\mathbf{u}\cdot \mathbf{v})\mathbf{u}=\mathbf{v}.$$Since $\Vert \mathbf{u}\Vert =1,$ $\mathbf{v}-(\mathbf{u}\cdot \mathbf{v})\mathbf{u}=\mathbf{v}.$ Then $$(\mathbf{u}\cdot \mathbf{v})\mathbf{u}=0.$$Again, since $\Vert \mathbf{u}\Vert =1,$ we must have $\mathbf{u}\cdot \mathbf{v}=0.$

Now, $$\begin{array}{rl}\mathbf{u}\cdot (\mathbf{v}\times \mathbf{w})& =\mathbf{u}\cdot (\mathbf{v}\times (\mathbf{u}\times \mathbf{v}+\mathbf{u}))\\ & =\mathbf{u}\cdot (\mathbf{v}\times (\mathbf{u}\times \mathbf{v})+\mathbf{v}\times \mathbf{u})\\ & =\mathbf{u}\cdot (\mathbf{v}\times (\mathbf{u}\times \mathbf{v}))+\mathbf{u}\cdot (\mathbf{v}\times \mathbf{u}).\end{array}$$By the vector triple product, $$\mathbf{v}\times (\mathbf{u}\times \mathbf{v})=(\mathbf{v}\cdot \mathbf{v})\mathbf{u}-(\mathbf{v}\cdot \mathbf{u})\mathbf{u}.$$Since $\Vert \mathbf{v}\Vert =1$ and $\mathbf{u}\cdot \mathbf{v}=0,$ this simplifies to $\mathbf{u}.$ Also, $\mathbf{u}$ is orthogonal to $\mathbf{v}\times \mathbf{u},$ so $$\mathbf{u}\cdot (\mathbf{v}\times \mathbf{w})=\mathbf{u}\cdot \mathbf{u}=\boxed{1}.$$