Is the tensor product a special case of the geometric product?

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I know the basics of geometric algebra and that the geometric product is the sum of the inner and outer products. I've also seen the tensor product described as an outer product, which makes sense. Tensors themselves are defined as members of a vector space that's constructed as the tensor product of two other vector spaces. That all leads to me think that the tensor product is just a special case of the geometric product, just like the dot product, cross product, wedge product, etc. However, when I did a web search about tensors in geometric algebra, I found multiple sources saying that geometric algebra can be embedded into tensor algebra, but not vice versa -- that all multivectors are tensors but not vice versa. But how can multivectors, which are just objects built from the geometric product, be a subset of tensors? Tensors themselves are objects built from the tensor product, a type of outer product, and outer products themselves are just half of the geometric product?

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Tensor product is not a type of outer product. You get the outer product by antisymmetrization of the tensor product. There is a universal property of the tensor product (of two spaces, say): any bilinear operation taking two vectors (one from each space) into the underlying field can be "modeled" on the tensor product (you can google universal property tensor product for details). Since the geometric product is bilinear, it can be factored through the canonical mapping $V\times W\to V\otimes W$ (namely, it is the composition of this mapping with a linear operator $V\otimes W\to Z$, where $Z$ is any vector space, in your case the whole algebra.

Geometric algebra includes the exterior algebra (antisymmetric tensors) but not all the tensors. The main point of geometric (Clifford) algebra is that it is an associative algebra, where elements have an inverse, etc.

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On the contrary, the geometric product can be viewed as a special case of the tensor product. The geometric product has the same rules as the tensor product, except for one extra: For a vector v: \begin{equation} v^2=|v|^2 \end{equation}