I will answer the question in the affirmative for $\mathfrak{sl}_n$, and in general provide a property of the Weyl group that would imply an affirmative answer in general. This property is true in type A, but not for the other rank 2 cases. The other rank 2 Lie algebras are small enough to be dealt with easily.

Let me assume for now that λ is regular and integral. Write $\lambda = x\cdot \lambda'$, where x is in the Weyl group and λ' is dominant. Write ch(L(λ)) as a rational function in reduced form, and let T be the set of positive roots α such that $(1-e^\alpha)$ does not appear in the denominator. Let G be the subgroup generated by all reflections in all elements of T. I claim that if g is in G, then gx≤x in Bruhat order.

Switch to indexing simples and Vermas by the Weyl group. We have
$$\operatorname{ch}(L(x))=\sum_y P_{x,y}\operatorname{ch}(M(y)).$$
The claim follows from two simple observations. Firstly, for $(1-e^\alpha)$ to not appear in the denominator, we must have $P_{x,y}+P_{x,sy}=0$, where s is the reflection for the root α. Secondly, that P_{x,x}=1 and P_{x,y}=0 unless y≤x.

Now we turn to the question. Consider two simples L(λ) and L(μ). From the above discussion we obtain two Weyl group elements x and y, together with two sets of positve roots T_{x} and T_{y}, and groups G_{x} and G_{y} generated by the corresponding reflections, so that $G_xx\leq x$ and $G_y y\leq y$. If $L(\lambda)\otimes L(\mu)$ is in category O, then $T_x\cup T_y$ is all positive roots by looking at the character. For L(λ) and L(μ) to be infinite dimensional, we must have that x and y are not the identity.

Now we have a condition on the Weyl group which is necessary for a negative answer. In the symmetric group, subgroups generated by reflections are products of smaller symmetric groups. It is easy to see that for any two proper such subgroups, there is a reflection which is not contained in either. Thus in type A, since every positive root appears in either T_{x} or T_{y}, either G_{x} or G_{y} is the entire symmetric group, which then contradicts the Bruhat ordering condition as x and y are not the identity.

If we move beyond regular or integral weights, then it becomes even harder to cancel factors in the denominator of the character, so the argument should be easier. Beyond type A, I haven't checked any non-type A Weyl groups of rank at least three to see if my criterion is enough to answer the question.