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.)