EDITED (my arithmetic being too hasty):
I assume you are working over a field like $\mathbb{C}$ of characteristic 0, where it's equivalent to work with the Lie algebra (and in any case $G$ might as well be taken to be semisimple).
A somewhat random exampleAt first sight I don't see an immediate reason for type $A_2$ seems to show how your proposed inclusion will typically failto be correct, though such an inclusion would presumably be "natural" when it exists. First Since the irreducible summands are uniquely determined in each tensor product, an inclusion would be the same as having the first list of summands included in the second list (multiplicity counted).
In any case it's helpful to clarify the role of dual representations here. First, recall that $V_\mu^*$ is itself irreducible of highest weight $\mu^*:= -w_\circ \mu$ (where $w_\circ$ is the longest element of the Weyl group). Moreover, on the right side you have $(\mu + \nu)^* = \mu^* + \nu^*$. In type $A_2$ Putting this together, your inclusion would be equivalent to a dominant weight abbreviated by $(r,s)$ with $r,s \in \mathbb{Z}^+$ has "dual" weightstatement that $(s,r)$. Now take$V_\lambda \otimes V_\mu$ is included in $\nu = (0,1)$ and consider the special case of your question: $$V_{(1,0)} \otimes V_{(2,0)} \hookrightarrow V_{(1,1)} \otimes V_{(1,2)} ?$$
Using Brauer's standard method$V_{\lambda+\nu} \otimes V_{\mu + \nu^*}$ for decomposing the tensor product on the right side, you don't get the irreducibleany dominant weight $V_{(3,0)}$ as a summand, which obviously occurs as a summand on the left side$\nu$.