Let $F$ be any field, and let $A$ be an associative unital finite dimensional $F$-algebra, with an $F$-linear involution $*$ (possibly trivial), that is an isomorphism of the $F$-vector space $A$ of order at most $2$, such that $(a_1a_2)^*=a_2^*a_1^*$ for all $a_i\in A$.
Define the unitary affine group scheme $U(A,*)$ by $$U(A,*)(R)=\{x\in A\otimes_FR\mid x^*x=1\},$$ for any commutative $F$-algebra $R$ (where we extend $*$ on $A\otimes_FR$ by tensoring with $Id_R$).
Let $J=Rad(A)$ be the Jacobson radical of $A$. Then $J^*=J$, since $1+AxA\subset A^\times$ if and only if the same holds for $x^*$ (apply $*$ to the inclusion). Therefore, $*$ induces an $F$-linear involution on $A/J$, and we may consider $U(A/J,*)$.
The projection $A\to A/J$ then induces a morphism $\pi:U(A,*)\to U(A/J,*)$.
Question 1. Is $\pi_{\bar{F}}:U(A,*)(\bar{F})\to U(A/J,*)(\bar{F})$ surjective, where $\bar{F}$ is an algebraic closure of $F$?
Question 2. If the answer to Question 1 is affirmative, is $\pi$ a surjective morphism of affine group schemes ,i.e. is the corresponding map $\pi^*$ between Hopf algebras injective ?
Remark 1. Let $f:G\to H$ be a morphism of affine group schemes. It is known that if $f$ is surjective, then $f_{\bar{F}}:G(\bar{F})\to H(\bar{F})$ is surjective. Moreover, if $H$ is smooth, the converse also holds.
Hence $U(A/J,*)$ is smooth, the two questions are equivalent.
However, $U(A/J,*)$ is not always smooth, even if $J=0$ and thus $A$ is semi simple.
For example, if $A=F$ and $*$ is the identity, then $U(A)=U(A/J)=\mu_2$, which is not smooth in characteristic two.
What I have obtained so far. Say that the involution $*$ is tame if there exists $a\in Z(A)$ such that $a^*+a=1$, and wild otherwise.
For example, if $\mathrm{char}(F)\neq 2$, any $F$-linear involution is tame, since we may take $a=\dfrac{1}{2}$.
Assuming that $*$ is tame, we have the following results:
(i) For any $R$, $\pi_R:U(A,*)(R)\to U(A/J,*)(R)$ is surjective.
(ii) The affine group scheme $U(A/J,*)$ is smooth.
In particular, Questions 1 and 2 have an affirmative answer.
Remark 2. Note however that $U(A/J,*)$ can be smooth even if $*$ is wild.
For example, if $A=M_2(F)$ and $*$ is defined by $M^*=HM^t H^{-1}$, where $H=\pmatrix{0 & 1\cr 1 & 0}$, then $U(A,*)=Sp_2=SL_2$, which is smooth. However, $*$ is wild if $\mathrm{char}(F)=2$, since $Z(A)=F I_2$ and $x^*+x=2x=0$ for all $x\in F.I_2$.
For the moment, I haven't found a counter example to Question 1 and I have no idea whether Questions 1 and 2 have a positive answer in the wild case, even if I suspect that the answer to both questions is NO.
Assume $F$ is algebraically closed. Then the map $A\to A/J$ splits and so there is a subalgebra $B$ of $A$ such that $A=B\oplus J$. This may not be a $\ast$-splitting, though, so if $b\in B$ we only know that $b^\ast=b_1+c$.
Your question 1 now becomes: given $b\in B$ with $b_1b=1$, can we find $x\in J$ with $(b+x)^\ast(b+x)=1$. Expanding out this gives $(b_1x+x^\ast b+cb)+(c+x^\ast x)x=0$. As a first approximation we want to find $x$ with $b_1x+x^\ast b=-cb$ mod $J^2$. Multiplying by $b$ then gives $x+bx^\ast b=-bcb$.
Note that $J/J^2$ is a $B$-bimodule, but $B$ is a product of full matrix rings over $F$, so $B\otimes B$ is again semisimple.
If we can solve this, then we can iterate and use that $J$ is nilpotent to answer question 1. I guess there are just a few cases to consider now. For example, either $B$ is just one full matrix ring with an involution, or else it as a product of two such and the involution swaps the two factors. Maybe this helps.