Is there a way to solve for angles and side lengths of this pentagon algebraically?
We are given the equations:
$$A=90^{\circ}, 2B+C=360^{\circ}, C+E=180^{\circ}, 2a=2c=d=e \text{ and say } a=1$$
Also we need to guarantee that the angles and sides actually fit together to give a pentagon.
e.g. $$b - a \cos(A) - c \cos(B) + d \cos(B+C) + e \cos(A+E)=0$$ and
$$a \sin(A) - c \sin(B) + d \sin(B+C) - e \sin(A+E) = 0$$
Also, squaring and adding the last two equations gives a nice symmetrical equation:
$$ \begin{split} 2(a b \cos (A)+b c \cos (B)+c d \cos (C)+d e \cos (D)+e a \cos (E))=a^2+b^2+c^2+d^2+e^2\\+2 (a c \cos (A+B)+b d \cos (B+C)+c e \cos (C+D)+d a \cos (D+E)+e b \cos (E+A)) \end{split}$$ which seems to be the cosine rule for pentagons or something.
Here's the solution with approximated angles. Thanks!
Notice first of all that $A+B+C+D+E=3\cdot180°$. Combining this with the other relations you give for the angles, we may rewrite them all as follows: $$ A=90°,\quad C=360°-2B,\quad D=270°-B,\quad E=2B-180°, $$ so that all angles can be expressed in terms of $B$ alone.
Your equation $$a \sin(A) - c \sin(B) + d \sin(B+C) - e \sin(A+E) = 0$$ can be then rewritten as $$1-3\sin(B)+2\cos(2B)=0$$ where I've also inserted the values $a=c=1$ and $d=e=2$. This equation is easy to solve and gives $$\sin(B)={\sqrt{57}-3\over8}, \quad\hbox{whence}\quad B=145.338°.$$ It is then easy to find the other angles and finally to obtain $b$ from the other equation: $$b =-\cos(B)-2\sin(2B)=2.6937$$