I think that there is indeed no such finite group $G$, whether simple or otherwise. Note first that the representation $5_{1}$ can be assumed to be faithful ( for if $K$ is its kernel, then the group $G/K$ has the same property), so from now on, we assume it faithful.

Note next that $Z(G) = 1$, since if $5_{1}$ lies over a linear character $\lambda$ of $Z(G)$, then we must have $\lambda^{2} =1 $ since the trivial character occurs in $5_{1} \otimes 5_{1}$. However $\lambda = \lambda^{2}$ since $5_{1}$ also occcurs in $5_{1} \otimes 5_{1}$. Hence $\lambda$ is trivial.

Now $ N = O_{5^{\prime}}(G)$ is Abelian by Clifford's Theorem. If $N$ is non-trivial, then it is also non-central, sincee $Z(G) = 1$, and it follwws from Clifford's Theorem that the representation $5_{1}$ is (up to equivalence) monomial. Then $G$ has an Abelian normal subgroup $A$ such that $G/A$ is isomorphic to  a subgroup of $S_{5}$. But in that case, $G$ has an Abelian normal Sylow $7$-subgroup, and $G$ has no irreducible character of degree $7$ (by a theorem of Ito, the degree of a complex irreducible character of $G$ divides $[G:A]$ whenever $A \lhd G$ is Abelian).
Hence it follows that $N = 1$. More generally, this argument shows that the representation $5_{1}$ is primitive, ie not (equivalent to one) induced from any proper subgroup of $G$.

There are several ways to finish from here. One is to invoke Brauer's classification of the finite primitive subgroups of ${\rm GL}(5, \mathbb{C})$ and note that none of these has tthe order of $G/Z(G)$ divisible by $7$. 

Another is to note that if $O_{5}(G)$ is non-trivial, then it is irreducibly represented by $5_{1}$, in which case $Z(G)$ has order divisible by $5$, a contradiction. 

Now we are reduced to the case $F(G) = 1$ and we continue until we see that $M = F^{\ast}(G)$ is a finite simple subgroup of ${\rm GL}(5,\mathbb{C})$ of orer divisible by $35$. But by a theorem of Feit, if $G$ is a finite simple subgroup (of order divisible by the prime $p$) of ${\rm GL}(p-2,\mathbb{C})$ for some prime $p$, then $p$ is a Fermat prime and $G = {\rm SL}(2,p-1)$ ( we may apply this with $p = 7$, so obtain a contradiction).