Up to irrelevant set-theoretic issues, every category $\mathbb{C}$ embeds as a full subcategory into:
complete and cocomplete cartesian closed category (in fact a Grothendieck topos); the embedding is given by the usual Yoneda functor $y \colon \mathbb{C} \rightarrow \mathbf{Set}^{\mathbb{C}^{op}}$; it preserves all existing limits (particularly all existing products), exponent and no non-trivial colimits,
complete and cocomplete co-cartesian closed category (in fact a co-topos); the embedding is given by the contravariant Yoneda functor on the opposite category $y^{op} \colon \mathbb{C} \rightarrow (\mathbf{Set}^{\mathbb{C}})^{op}$; it preserves all existing colimits (particularly all existing coproducts), co-exponents and no non-trivial limits,
complete and cocomplete category with the embedding preserving all existing limits and colimits (particularly all existing products and coproducts); the embedding is given by the Dedekind-MacNeille completion $\mathit{dm} \colon \mathbb{C} \rightarrow \mathit{DM}(\mathbb{C})$; see the excellent answer provided by Todd Trimble to my old question.
One may also show that we cannot have three at once: there is no universal embedding to a complete and cocomplete category with (co)exponents that preserves all existing limits, colimits and (co)exponents.
So, to directly address your question: as long as you take only limits and colimits (products and coproducts) into consideration, you may always enlarge your category in such a way that all limits and colimits will exist. However such enlargement generally will not preserve all important properties of your structures.
Perhaps, a better way to look at the issue is that in any category all products and coproducts "externally" exist, but sometimes we do not have enough objects to "internalize" them in the category: i.e. there is always a "product distributor" $\hom(\Delta(-), X_i)$ (where $\Delta$ is the diagonal functor), but there may not exist a "product object" that represents the distributor as an embedded functor $\hom(-, \prod_i X_i)$.
For fields the most basic problem is fields of different characteristic. If $K$ and $L$ are fields, then the product $K\times L$ (if it existed) would have to be a field that can map into both $K$ and $L$, which means in particular it has the same characteristic as both $K$ and $L$, which is impossible if $K$ and $L$ have different characteristic.
Even if you restrict to fields of a fixed characteristic, however, you still don't have products. For instance, in the category of fields of some fixed characteristic, let $K$ be any field with a non-identity endomorphism $e:K\to K$. Then I claim there is no product $K\times K$. For suppose there were; call it $P$, and let $p:P\to K$ and $q:P\to K$ be the two projections. Considering $L=K$ and the identity maps $f=g=1:L\to K$, by the universal property of the product there must exist $h:L\to P$ such that $ph=f$ and $qh=g$. Since $f$ and $g$ are the identity and every map of fields is injective, this means that $h$ must be an isomorphism and $p=q$ is its inverse. But now consider $g'=e:L\to K$; there is then an $h':L\to K$ such that $ph'=f$ and $qh'=g'$. Since $p=q$, this means $f=g'$, which is a contradiction since we assumed $e\neq 1$.
The case of smooth manifolds with boundary is fairly subtle (indeed, the category of smooth manifolds has some limits you might not expect it to have). Let me sketch a proof that there is no product $[0,1)\times[0,1)$ in the category of smooth manifolds with boundary. Suppose you had such a product $P$. By considering maps from the $1$-point manifold to $P$, you can identify the underlying set of $P$ with the cartesian product $[0,1)\times[0,1)$. By considering smooth maps from $\mathbb{R}$, you can show that $P$ is connected. By considering smooth maps from $\mathbb{R}^2$, you can show that $P$ is $2$-dimensional and that $\{0\}\times [0,1)\cup[0,1)\times\{0\}$ must be its boundary (here we use that a point is in the interior of a $2$-manifold iff there is an injective smooth map from $\mathbb{R}^2$ to the manifold whose image contains the point but no such map from $\mathbb{R}^n$ for any $n>2$). Again using smooth maps from $\mathbb{R}$, you can show that this boundary must be connected, and that (via the classification of $1$-manifolds) in fact it must be diffeomorphic to $\mathbb{R}$. Such a diffeomorphism $\mathbb{R}\to P$ onto the boundary then gives a smooth injection $i:\mathbb{R}\to\mathbb{R}^2$ whose image is $\{0\}\times [0,1)\cup[0,1)\times\{0\}$, and $i$ has the property that every smooth map $j:\mathbb{R}\to\mathbb{R}^2$ whose image is contained in $\{0\}\times [0,1)\cup[0,1)\times\{0\}$ factors smoothly through $i$. You can show that no such $i$ exists by examining what it does near the corner: assuming without loss of generality that $i(0)=(0,0)$, then all the derivatives of $i$ must vanish at $0$. But then $j(x)=i(\sqrt[3]{x})$ is smooth and does not factor through $i$, which is a contradiction.
For other examples of categories without products, it is very easy to get examples if you look at small categories you think of as an individual algebraic structure, rather than categories like "the category of all structures of a certain kind". For instance, you can form a category with any set of objects and no non-identity morphisms, and this doesn't have products of distinct objects. Or you can take a monoid and consider it as a category with one object, and it will rarely have a product of two copies of the object.
Best Answer
Sure. For instance, take the category of sets whose cardinality is not $4$. This category obviously has all products except for a product of two $2$-element sets (or products of higher arity where two of the factors have two elements and the rest have one, which are essentially the same thing since a singleton is terminal). But a weak product of two $2$-element sets (call them $A$ and $B$) does still exist. For instance, let $C$ be any set with more than one element and consider $P=A\times B\times C$ with its projections $p$ and $q$ to $A$ and $B$. I claim $(P,p,q)$ is a weak product of $A$ and $B$ (that is, it satisfies the definition of a product except for uniqueness of the maps). Indeed, let $Q$ be any set and $f:Q\to A$, $g:Q\to B$. Pick any function $c:Q\to C$ (such a function exists since $C$ is nonempty), and define $h:Q\to P$ by $h(x)=(f(x),g(x),c(x))$. Then $ph=f$ and $qh=g$, as desired.
(Note that no product of $A$ and $B$ exists in this category, since by considering maps from a singleton set, such a product would need to have $4$ elements.)